Question

An object is dropped from a​ tower, 400 ft above the ground. The​ object's height above ground x seconds after the fall is ​s(x)equals400minus16xsquared. About how long does it take the object to hit the​ ground? What is the​ object's velocity at the moment of​ impact?

Answer 1

1) 5 s

The vertical position of the object is given by

`y(t) = h - (1)/(2)gt^2 =  400 - 16 t^2`

where

h=400 ft represents the initial height

g = 32 ft/s^2 is the acceleration of gravity

t is the time

We want to find the time t at which the object reaches the ground, so the time t at which

y(t) = 0

By substituting this into the equation, we find

`0 = 400 - 16t^2\\t=\sqrt{(400)/(16)}=5 s`

2) 160 ft/s

The object is released from rest, so the initial velocity is zero

u = 0

The final vertical velocity can be found by using

`v^2 - u^2 = 2ah`

where

v is the final velocity

a = 32 ft/s^2 is the acceleration of gravity

h = 400 ft is the vertical distance covered

Solving for v, we find

`v=√(u^2 +2ay)=√(2(32 ft/s)(400 ft))=160 ft/s`