Question

A ball is dropped from the top of a building that is known to be 400 feet high. The formula for finding the height of the ball at any time is h=400−16t2, where t is measured in seconds.
How many seconds did it take for the ball to reach a height of 256 feet above the ground?

Answer 1

Answer:

3 seconds

Explanation:

Height of the building = 400 feet

Height of the ball from the ground is given by

h=400−16t²

This formula has been derived from

$s=ut+\frac{1}{2}at^2$

a = Acceleration due to gravity = 32 ft/s²

u = Initial velocity = 0

t = Time taken

Substituting all the values we get

$s=0t+\frac{1}{2}32t^2\\\Rightarrow s=16t^2$

This is the height of the ball from the top of the building

The height of the ball from the ground will be

h = 400-s

⇒h = 400−16t²

When h = 256 ft

$256=400-16t^2\\\Rightarrow t=\sqrt{\frac{256-400}{-16}}\\\Rightarrow t=3\ s$

Time taken by the ball to reach a height of 256 feet above the ground is 3 seconds