Download H(t)

Denise decided (for reasons unknown) to toss an apple into the air. She gave a little
jump and thrust the apple into the air, releasing the apple from a height of 10 feet
above the ground with an initial velocity of 40 feet per second. (See solutions to
problems 1 and 4 attached.)
Once released, the apple went upward with velocity decreasing over time.
(According to my solution to problem 3, velocity decreases at a constant rate of
–32 feet per second per second.) The apple reached its maximum height of 35 feet
1.25 seconds after it was released. I know this was the maximum height because at
that point in time the velocity changed from being positive to being negative, which
means that the height above the ground changed from increasing to decreasing.
(See solution to problem 5 attached.)
Curious to know when the apple was 30 feet above the ground, I solved the
equation
H (t) = −16t 2 + 40t + 10 = 30, or − 16t 2 + 40t − 20 = 0,
or − 4t 2 + 10t − 20 = 0
and got the solutions t=0.427 and t=2.927 , meaning that about 0.427 seconds
after the apple was released it was 30 feet above the ground and going up, while at
about 2.927 seconds after the apple was released it was again 30 feet above the
ground and on its way down.
Wondering if the apple’s speed was ever as much as the initial 40 feet per second,
and knowing that the velocity was decreasing, I solved the equation v(t)=-40,
getting t=2.5, meaning that 2.5 seconds after the apple was released it was again
traveling at a speed of 40 feet per second (and velocity –40 feet per second). I
was pleased by the symmetry of having this same speed as the initial speed
reached in exactly double the time it took to reach its maximum height. Then I
noticed that height was once again 10 feet above the ground at this same point in
time, another symmetry.
The apple finally hit the ground about 2.729 seconds after its release from
Denise’s hand. (See the solution to problem 2.)
Problem Solutions.
1. Since
H (t) = −16t 2 + 40t + 10 , H (0) = 10 , which means that the
apple is 10 feet above the ground when it is released.
2. The apple hits the ground at a time t for which
H (t) = −16t 2 + 40t + 10 = 0 . Equivalently, −8t 2 + 20t + 5 = 0
Using the quadratic formula on the above reduced equation, we get two
−20 + 20 2 − 4(−8)(5)
solutions: t =
and
2(−8)
−20 − 20 2 − 4(−8)(5)
t=
. These two solutions reduce to
2(−8)
(35)
t = 1.25 −
≈ 1.25 − 1.479 = −0.229
4
(35)
and t = 1.25 +
≈ 1.25 + 1.479 = 2.729 . (Please see the end of
4
this document for details.) The negative solution refers to a time before Denise
tosses her apple, so it can be discarded. So, the apple hits the ground about
2.729 seconds after it was tossed into the air.
3. To find H’(t), look at
H (t + h) − H (t) = −16(t + h)2 + 40(t + h) + 10 − (−16t 2 + 40t + 10)
= −16t 2 − 32ht − 16h 2 + 40t + 40h + 10 + 16t 2 − 40t − 10
= −32ht − 16h 2 + 40h
Then
H (t + h) − H (t) −32ht − 16h 2 + 40h h(−32t − 16h + 40)
=
=
h
h
h
= −32t − 16h + 40, if h ≠ 0
and finally, v(t) = H '(t) = lim (−32t − 16h + 40) = −32t + 40 .
h→0
The units for v(t) are feet per second.
4. Since v(0)=40, the apple is traveling upwards at 40 feet per second at t=0.
Similarly, v(1)=-32+40=8, so the apple is traveling upwards at 8 feet per
second at t=1, and since v(2)=-64+40=-24, the apple is traveling downwards
at 24 feet per second at t=2. (The velocity at t=2 is –24 feet per second.)
5. To determine whether the velocity is ever zero, suppose that there is such a
time t for which v(t)=-32t+40=0 and solve for t. The equation holds if t=1.25.
At this instant, the velocity is changing from positive to negative and the
apple is changing from going up to going down.
6. Since the apple is changing from going up to going down when t=1.25, the
maximum height the apple reaches occurs when t=1.25. Since
5
5
5
H ( ) = −16( )2 + 40( ) + 10 = −25 + 50 + 10 = 35 , the
4
4
4
maximum height reached is 35 feet.
−20 − 20 2 − 4(−8)(5)
Details of the simplification of t =
:
2(−8)
t=
=
−20 + 560 −20 + (16)(35) −20 + 4 (35)
=
=
2(−8)
2(−8)
2(−8)
5 − (35)
4