Download Resultant velocity of a horizontal projectile

De LaSalle-Araneta University
Department of Technology
Final Product Requirement
Engineering Mechanics
Portfolio Compilation of Problems and Solutions
Relating to Engineering Mechanics
Submitted to: Engr. Aga Madelo
Submitted by: Peter Glenn Nuestro
Curvilinear motion
and the acceleration from the velocity function (or displacement function), using:
These formulae are only appropriate for rectilinear motion (i.e. velocity and acceleration in a straight line). This is
inadequate for most real situations, so we introduce here the concept of curvilinear motion, where an object is
moving in a plane along a specified curved path.
We generally express the x and y components of the motion as functions of time. This form is called parametric
form. (See another example using parametric form in Lissajous Figures.)
Example 1: Parametric Equations
Draw the curve
y(t) = cos t,
x(t) = sin t
for t = 0 to 2π in 0.5 intervals.
First, we need to set up a table of values which we obtain by substituting various values of t:
t
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
x(t) 0 .48 .84 1.0 .91 .60 .14 -.35 -.76 -.98 -.96 -.71 -.28
y(t) 1 .88 .54 .07 -.42 -.80 -.99 -.94 -.65 -.21 .28 .71 .96
Answer
We plot our 13 points, starting at (1, 0) as follows:
We see that we have formed a circle, centre (0,0), radius 1 unit.
Notice that the variable t does not appear in this graph, just the variables x and y.
Horizontal and Vertical Components of Velocity
The horizontal component of the velocity is written:
and the vertical component is written:
We want to find the magnitude of the resultant velocity v once we know the horizontal and vertical components. We
use:
The direction θ that the object is moving in, is found using:
Example 2
If x = 5t3 and y = 4t2 at time t, find the magnitude and direction of the velocity when t = 10.
Answer
When t = 10, the particle is at (5000, 400).
Here is the graph of the motion.
Note:


The axes are x and y (and do not involve t).
The particle is accelerating as time goes on (the red dots are at one second intervals)
We are told that
x = 5t3
So
At t = 10, the velocity in the x-direction is given by:
Also, y = 4t2 so the velocity in the y-direction is:
When t = 10, the velocity in the y-direction is:
So the magnitude of the velocity will be:
Now for the direction of the velocity (it is an angle, relative to the positive x-axis):
So θv = 0.053 radians = 3.05°.
Example 3
If
and
at time t, find the magnitude and direction of the velocity when t = 2. Plot the curve.
Answer
When t = 2, the particle is at (8, 0.6).
so
At t = 2,
dx/dt = vx = 20/25 = 0.8 ms-1.
Also,
When t = 2,
dy/dt = vy = 0.5 ms-1.
So
so
Now for the direction:
So θv = 0.558 radians
Acceleration of a Body in Curvilinear Motion
The expressions for acceleration are very similar to those for velocity:
Horizontal component of acceleration:
Vertical component of acceleration:
Magnitude of acceleration:
Direction of acceleration:
Example 4
A car on a test track goes into a turn described by x = 0.2t3, y = 20t− 2t2, where x and y are measured in metres and t
in seconds.
(i) Sketch the curve for 0 ≤ t ≤ 60.
(ii) Find the acceleration of the car at t = 3.0 seconds.
Answer
(ii) Acceleration:
Horizontal acceleration:
x = 0.2t3
At t = 3.0, ax = 3.6
Vertical acceleration:
y = 20t − 2t2
At t = 3.0, ay = -4
Now
and
So the car's acceleration has magnitude 5.38 ms -2, and direction 312° from the positive x-axis.
What if x and y are NOT given as functions of t?
Example 5
A particle moves along the path y = x2 + 4x + 2 where units are in centimetres. If the horizontal velocity vx is constant
at 3 cm s-1, find the magnitude and direction of the velocity of the particle at the point (-1, -1).
Answer
This is a different situation to the other examples. This time we have y in terms of x, and there are no expressions
given in terms of "t" at all.
To be able to find magnitude and direction of velocity, we will need to know
and
But the question already gives us
so all we need to find is
.
To find this, we differentiate the given function with respect to t throughout using the techniques we learned back in
implicit differentiation:
y = x2 + 4x + 2
Since
and we want to know the velocity at x = -1, we substitute these two values and get:
So now we have vy = 6 cm s-1.
So the magnitude of the velocity is given by:
The direction of the velocity is given by:
So the velocity is 6.7 cm s-1, in the direction 63.4°.
Example 6
A rocket follows a path given by (distances in km):
If the horizontal velocity is given by V(x) = x, find the magnitude and direction of the velocity when the rocket hits the
ground (assume level terrain) if time is in minutes.
Answer
Let's first see a graph of the motion, to better understand what is going on.
We can see that the rocket hits the ground again somewhere around x = 9.5 km. At this point, the horizontal velocity
is positive (the rocket is going left to right) and the vertical velocity is negative (the rocket is going down).
"V(x) = x" means that as x increases, the horizontal velocity also increases with the same number (different units, of
course). So for example, at x = 2 km, the horizontal speed is 2 km/min, and at x = 7 km, the horizontal speed is 7
km/min, and so on.
To calculate the magnitude of the velocity when the rocket hits the ground, we need to know the vertical and
horizontal components of the velocity at that point.
(1) Horizontal velocity. We just need to solve the following equation to find the exact point the rocket hits the
ground:
Factorising gives:
And solving for 0 gives us x = 0, x = -3√10, x = 3√10
We only need the last value, x = 3√10 ≈ 9.4868 km (This value is consistent with the graph above).
So the horizontal speed when the rocket hits the ground is 9.4868 km/min (since V(x) = x).
(2) Vertical velocity. We now need to use implicit differentiation with respect to t (not x!) to find the vertical
velocity.
But we already know dx/dt and x at impact, so we simply substitute:
This gives us a negative velocity, as we expected before:
So now we need to find the magnitude of the velocity. This takes into account both the horizontal and vertical
components.
Magnitude =
Substituting, we have:
Velocity has magnitude and direction. Now for the direction part.
Angle of motion:
Substituting our vertical and horizontal components, we have:
In degrees, this is equivalent to
-1.107148718 × 57.25578 = -63.3907°
We can see that this answer is reasonable by zooming in on the portion of the graph where the rocket hits the ground
(with equal-axis scaling):
So in summary, the velocity of the rocket when it hits the ground is 21.2 km/min in the direction 63.4° below the
horizontal.
Horizontal motion
1. A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for
the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
Given :
x = ???
vix = 2.4 m/s
ax = 0 m/s/s
y = -0.60 m
viy = 0 m/s
ay = -9.8 m/s/s
-0.60 m = (0 m/s)•t + 0.5•(-9.8 m/s/s)•t2
-0.60 m = (-4.9 m/s/s)•t2
0.122 s2 = t2
t = 0.350 s (rounded from 0.3499 s)
x = (2.4 m/s)•(0.3499 s) + 0.5•(0 m/s/s)•(0.3499 s)2
x = (2.4 m/s)•(0.3499 s)
x = 0.84 m (rounded from 0.8398 m)
2. A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of
the hill. Determine the initial horizontal velocity of the soccer ball.
Given:
x = 35.0 m
y = -22.0 m
vix = ???
viy = 0 m/s
ax = 0 m/s/s
ay = -9.8 m/s/s
Use y = viy • t + 0.5 • ay • t2 to solve for time; the time of flight is 2.12 seconds.
Now use x = vix • t + 0.5 • ax • t2 to solve for vix
Note that ax is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12
s for t, the vix can be found to be 16.5 m/s.
8. A ball is thrown into the air with an initial vertical velocity of 5.0m/s in addition to its horizontal velocity of 25.0m/s,
how far would the ball travel horizontally before it reached it's peak?
vf = vi + at
I've also tried vf = 1/2g +at
0 = 5.0m/s + (-9.8m/s^2)t
I put the vi for the inital velocity for vertical, then I inputed -9.8 because of the velocity being countered by the
gravitational pull.
then I rearange to -9.8m/s^2 by its self by doing
0 + (9.8m/s^2)t = 5.0m/s
then i divide both sides by 9.8m/s^2, looking like this
\frac{9.8m/s^2}{9.8m/s^2} = \frac{5.0m/s}{9.8m/s^2}
which would then leave me with t = .51 seconds
with that i would multiply that by the horizontal distance, looking like
.51 seconds x 25m/s = 12.8m
since 12m is the distance, i divide that by 2, giving me 6.4m reaching its peak.
3. A ski jumper leaves the ski track in the horizontal direction with a speed of 25 m/s. The landing incline below him
falls off with a slope of 35 degrees. Where does he land on the incline?
x = v0*t = (25)t
y = (0)t + (1/2)at^22 = - (1/2)(9.8)t^2
x = d*cos(35)
y = d*sin(35)
Setting the equations together d = 109
x = (109)cos(35) = 89.3
y = -(109)sin(35) = -62.5
y = v0*t - (1/2)gt^2
y- y0 = v0*t - (1/2)gt^2
y = y0 - (0)t - (1/2)gt^2
y = (v0 sin(theta))/(v0 cos(theta)) - (1/2)g(x/25)^2
y = tan(theta) - (1/2)g(x^2/ 625)
Projectile motion
Projectile motion is an example of curved motion with constant acceleration. This is the two dimensional motion of a
particle thrown obliquely into the air. The ideal motion of a cricket ball, a golf ball or a bullet is an example of projectile
motion. We assume that the effect air could have on their motion is negligible. Moreover, we also neglect

The effect due to curvature of the Earth

The effect due to rotation of the Earth
For all points on the trajectory
The acceleration due to gravity 'g' is constant in magnitude and direction.
Types of projectiles
Horizontal projectile
If a body is projected horizontally from a certain height with a certain velocity, then the body is called a horizontal
projectile.
Oblique projectile
If a body is projected at a certain angle with the horizontal, then the body is called an oblique projectile. The motion of
a projectile is a two dimensional motion. So it can be discussed in two parts

Horizontal motion

Vertical motion
These two motions take place independent of each other. This is called the principle of physical independence of
motion.
At any instant, the velocity of a projectile has two components

Horizontal component

Vertical component.
The horizontal component remains unchanged throughout the flight. The vertical component is continuously affected
by the force of gravity. Therefore, while the horizontal motion is a uniform motion, the vertical motion is a uniformly
accelerated motion.
Horizontal projectile
The figure above illustrates a body thrown horizontally from a point O with a velocity
The point O is at a certain
height above the ground.� Let x and y be the horizontal and vertical distances covered by the projectile,
respectively, in time t. Therefore, at time t, the projectile is at p.
In order to calculate x, let us consider the horizontal motion, which is uniform motion. This is because the only force
acting on the projectile is the force of gravity. This force acts vertically downwards and hence, the horizontal
component in zero. Therefore, the equations of motion of the projectile for the horizontal direction is just the equation
of uniform motion in a straight line.
x = vt ------------------ (i)
In order to calculate y, the vertical motion of the projectile is considered. Since the vertical motion is controlled by the
force of gravity, it is an accelerated motion. The initial velocity, vy (0), in the vertically downward direction is zero.
Since the Y-axis in the figure above is taken downwards, the downward direction is regarded as the positive direction.
So, the acceleration of the projectile is + g.
From the equation
Here vy (0) is taken as zero because both distance and time are being measured from the origin O.
From equation (1)
Substituting for t from the above equation in equation (2) we have,
is a constant for a projectile projected upwards with a definite velocity v and at a place with a definite value of 'g'.
Equation (3) is a second-degree equation in x, a first-degree equation in y and is the equation of a parabola.
Therefore, a body thrown horizontally from a certain height above the ground follows a parabolic trajectory till it hits
the ground.
Resultant velocity of a horizontal projectile:
In this section, let us calculate the resultant velocity of the projectile
of time t. Vx and Vy are the horizontal and vertical components of
The magnitude of the resultant velocity
is given by,
, at any point p on the trajectory, in an interval
as illustrated in the figure below.
Tension
1. A box weighing 70 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is
hung from the other end
Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs each
of the following.
35N is the weight.
Solution:
T - 35 = 0 (T is the tension in the rope) so T =35.
T is positive because its pulling up, and the weight is negative because its pushing down.
To find the normal force, you need to know that it is always perpendicular to the contact surface pushing on the
object. So in this case the normal force is acting straight up on the box.
So looking at the box, assuming that the pulley is frictionless, the sum of the forces in the Y is 0 so
N + T - 70 = 0 (N is the normal force, what you want to find)
so substituting T = 35 into the equation
N + 35 - 70 = 0
so
N = 35
2. An elevator with a mass of 2840 kg is given an upward acceleration of 1.22 m/s^2 by a cable. (a) Calculate tension
in the cable. (B) what is the tension when the elevator is slowing at the rate of 1.22 m/s^2 but is still moving upward.
F = ma
F = (2840)(9.8+1.22)
F = 31.3kN
F = ma
F = (2840)(9.8-1.22)
F = 24.4 kN
3. A 20 kg loudspeaker is suspended 2 meters below the ceiling by two 3 meter long cables that angle outward at
equal angles. What is the tension in the cables? The answer is 147 N anyone know how this is found?
Let the tension of cable be T and the the angle that each cable makes with vertical be 'theta'. cos theta = 2/3. T is
given by
2T cos theta = 20*9.81 or T = (20*9.83)/[2*(2/3)] = 147 N
4. Let the tension of cable be T and the the angle that each cable makes with vertical be 'theta'. cos theta = 2/3. T is
given by
2T cos theta = 20*9.81 or T = (20*9.83)/[2*(2/3)] = 147 N
sin(θ) = opposite/hypotenuse
sin(θ) = 634/680
= .9323529412
arcsin(.9323529412) = θ
69° = θ_1
Now for the second angle, we just use cos(θ) instead:
cos(θ) = adjacent/hypotenuse
cos(θ) = 634/680
= .9323529412
arcos(.9323529412) = θ
21° = θ
5. The distance between two telephone poles is 47.0 m. When a 0.500 kg bird lands on the telephone wire midway
between the poles, the wire sags 0.210 m.
How much tension does the bird produce in the wire? Ignore the weight of the wire.
47/2 = 23.5m = midpoint
midpoint = 23.5m
tetha (measured from horizontal line)
tan (tetha) = 0.21/23.5
sin (tetha) = 8.51*10^(-3)
cos (tetha) = 1
ΣFy = 0
2T sin (tetha) - W_bird = 0
T = W_bird/(2 sin (tetha))
T = 0.5 * 9.8/(2 * 8.51*10^(-3))
T = 287.897 N
6. A 279 kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined
at 30.0° with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.820, and
the log has an acceleration of 0.900 m/s2. Find the tension in the rope.
Formulae:
Ff = coefficient of friction X Fn
Weight of log = 279g kg
Horizontal component of weight (parallel to the surface of the ramp):
sin 30 X 279g
Vertical component of weight (perpendicular to surface of ramp):
cos 30 X 279g
Frictional force = cos 30 X 279g X 0.82
Total force opposite to the direction of applied force =
cos 30 X 279g X 0.82 + sin 30 X 279g
The log has a net acceleration of 0.9, ie net force applied =
0.9 X sin 30 X 279g
Tension in the rope - forces in opposite direction = net force applied
Tension in the rope
= net force applied + forces in opposite direction
= 0.9 X sin 30 X 279g + cos 30 X 279g X 0.82 + sin 30 X 279g
= 4543.79N
7. You pull downward with a force of 26.7 N on a rope that passes over a disk-shaped pulley of mass 1.14 kg and
radius 0.0723 m. The other end of the rope is attached to a 0.608-kg mass. Calculate the tension in the rope on both
sides of the pulley. Enter tension for the part of the rope that you are pulling on first. Then enter the tension for the
part of the rope with the mass.
T1 - 1.14 (9.8) = 1.14 a
26.7 (0.0723) - T1 (0.0723) = 1/2 (1.14) (0.0723)^2 alpha
alpha = a / (0.0723)
three equations with three unknowns, solve for T1.
T2 = 26.7
Edit: I made a mistake in the last equation, it is
alpha = a / (0.0723) and I had it as alpha = a (0.0723)
Dynamic Concept on Plane motion
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