Question

HELP QUICK - An object is dropped from a​ tower, 159 ft above the ground. The​ object's height above ground t sec into the fall is s =159−16t^2. a. What is the​ object's velocity,​ speed, and acceleration at time​ t? b. About how long does it take the object to hit the​ ground? c. What is the​ object's velocity at the moment of​ impact?

Answer 1

The velocity is the rate of change of the object's position:


`s(t)=159-16t^2\implies v(t)=s'(t)=-32t`

Acceleration is the rate of change of velocity:


`a(t)=v'(t)=s''(t)=-32`

The time it takes for the ball to hit the ground is the time
`t` for which
`s(t)=0`, since
`s(t)` describes the position of the ball above the ground.


`159-16t^2=0\implies 159=16t^2\implies t=\sqrt{(159)/(16)}\approx3.15`

(in seconds)

The object's velocity at this time can be found by plugging in this value of
`t` in to the velocity function.


`v(3.15)\approx-32(3.15)\approx100.88`

(in feet per second)