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Math 114 – Rimmer
13.2 Projectile Motion
A projectile is launched, we want to find the parametric equations for the path.
v0
the projectile has mass m
assume that gravity is the only force acting on the
projectile after it is launched ( neglect air resistance )
r0
Force due to gravity : F = ( −mg ) j
a = (−g ) j
Newton's Second Law : F = ma
v = ∫ a ( t ) dt
r = ∫ v ( t ) dt
v (t ) =
r (t ) =
⇒ ( −mg ) j = ma
Math 114 – Rimmer
13.2 Projectile Motion
a = (−g ) j
Often you are given an initial height, an initial speed,
v ( t ) = − gtj + v 0
and the angle θ at which the projectile is launched.
1 2
r ( t ) = − gt j + v 0t + r0
2
initial speed
v 0 = v 0 cos θ , v 0 sin θ
v0
r0 = 0, h
a = (−g ) j
a ( t ) = 0, − g
v ( t ) = 0, − gt + v 0 cos θ , v 0 sin θ
v (t ) =
θ
h=
r0
1
r ( t ) = 0, − gt 2 + v 0 cos θ , v 0 sin θ t + 0, h
2
r (t ) =
Math 114 – Rimmer
13.2 Projectile Motion
A baseball is hit 3 feet above the ground level at 100 feet per second
and at an angle of 45 with respect to the ground. Find the maximum
h=
v0 =
height reached by the baseball. Will it clear a 10 foot fence located
300 feet from home plate?
r (t ) =
(v
0
cos θ ) t , h + ( v 0 sin θ ) t −
1 2
gt
2
= (100 cos 45° ) t ,3 + (100sin 45° ) t −
θ =
g =
1
( 32 ) t 2
2
r (t ) =
The maximum height occurs when the __________ of the velocity vector is _____.
v ( t ) = 50 2,50 2 − 32t
⇒ 50 2 − 32t = 0 ⇒ t =
25 2
≈ 2.21 sec.
16
 25 2 
 25 2 
The maximum height = 3 + 50 2 
−
16



16
16




2
The ball is 300 ft. from home plate when the ___________ of the _______ vector is 300.
⇒ 50 2t = 300
6
t=
= 3 2 ≈ 4.24 sec.
2
Plug this time into the ___________ of the
position vector to find the height of the ball.
(
)
(
The height of the ball at the fence = 3 + 50 2 3 2 − 16 3 2
)
2
Math 114 – Rimmer
13.2 Projectile Motion
Math 114 – Rimmer
13.2 Projectile Motion
a=
v (t ) =
r (t ) =
Math 114 – Rimmer
13.2 Projectile Motion