Page 48 - How To Solve Word Problems In Calculus

t = 6, v =−32(6) + 256 = 64 ft/sec. The height of the pro-
2
jectile when t = 6is h =−16(6) + 256(6) = 960 ft.
dh
(c) v = =−32t + 256. When t = 10, v =−64 ft/sec. The
dt
negative velocity indicates that the projectile is moving in
the negative direction (downward) at a speed of 64 ft/sec.
2
The height of the projectile when t = 10 is h =−16(10) +
256(10) = 960 ft.
(d) v =−32t + 256
0 =−32t + 256 ← Maximum height is reached
when v = 0
32t = 256
t = 8 sec
2
h max =−16(8) + 256(8) ← Maximum height is reached
after 8 sec
= 1024 feet
(e) The projectile returns to the ground when h = 0.

2
h(t) =−16t + 256t
2
0 =−16t + 256t
0 =−16t(t − 16)
t = 0 t = 16

Since the ball was thrown at time t = 0, the ball returns to
the ground after 16 seconds. Its velocity at this time, from
the velocity function v(t) =−32t + 256 is

v (16) =−32(16) + 256 =−256 ft/sec

The negative velocity indicates that the ball is travel-
ing downward at a speed of 256 ft/sec when it hits the
ground.
(f ) When t = 0, h = 0 (the projectile is at ground level) and
v = 256 ft/sec. This is the initial velocity.As t increases,
the height h increases and the velocity v decreases, reach-
ing a value of 0 when t = 8 sec. At this point the max-
imum height of the projectile is reached, 1024 ft. As
t increases past 8 sec, v becomes negative and h gets

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