A projectile is fired vertically upward and is s feet above the ground t seconds after being fired, where s=256t — 16t^2. Find (a) the speed of the projectile 4 seconds after being fired, Complete question is in the image ?

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Answer 1 · 2017-09-03T08:19:45

#s = 256t — 16t^2#

Velocity => ds/dt = 256 - 32t

Hence, after 4 seconds we have:

ds/dt = 256 - 32(4) => 128 feet/sec

Maximum height is acheived when ds/dt = 0

i.e. when 256 - 32t = 0

=> t = 256/32 = 8 seconds

so, maximum height is #s = 256(8) — 16(8)^2#

i.e. #s = 2048 - 1024 => 1024 feet#

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