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

$y(t) = h - \frac{1}{2}gt^2 = 400 - 16 t^2$

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

$0 = 400 - 16t^2\\t=\sqrt{\frac{400}{16}}=5 s$

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

$v=\sqrt{u^2 +2ay}=\sqrt{2(32 ft/s)(400 ft)}=160 ft/s$