A baseball thrown directly upward at 96 ft/sec has velocity V(t) = 96-32t ft/sec

Question

A baseball thrown directly upward at 96 ft/sec has velocity V(t) = 96-32t ft/sec at time t seconds.

(b) When does the baseball reach the peak of its flight? How high does it go?

the peak is reached after __________ sec?

The max height is ___________ ft?

_________ft?

A baseball thrown directly upward at 96 ft/sec has velocity V(t) = 96-32t ft/sec at time t seconds. (b) When does the baseball reach the peak of its flight? How high does it go? the peak is reached after __________ sec? The max height is ___________ ft? (c) How high is the baseball at time A baseball thrown directly upward at 96 ft/sec has ?

Explanation / Answer

At the maximum height, the velocity will be zero, so differentiate and find t when v(t) = 0...

s(t) = 96t - 16t^2
v(t) = 96 - 32t = 0
t = 3'

Find the height at this time...

s = (96)(3) - (16)(9)
s = 144 feet.

For (b), find the time it takes to reach that height...

128 = 96t - 16t^2
t^2 - 6t + 8 = 0
(t - 2)(t - 4) = 0
t = 2 and t = 4

We're interested in t = 4, since t = 2 is when the ball is travelling up, so plug this time into velocity...

v(t) = 96 - 32(4)
v(t) = -32 ft/s, or 32 ft/s down

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