Question

An object is dropped from a height H above the ground. This free-falling object requires 1.02 s to travel the last 39 m before hitting the ground. From what height H above the ground did the object fall?

Answer 1

Answer:

H = 95.3 m

Explanation:

Let the height from which object fall down is H

so the total time to fall is given as

$H = \frac{1}{2}gt^2$

now it cover 39 m in last 1.02 s of its path

$H - 39 = \frac{1}{2}g(t - 1.02)^2$

so now we have

$\frac{1}{2}gt^2 - 39 = \frac{1}{2}g(t - 1.02)^2$

by solving above equation

$t = 4.41 s$

Now the total height is given as

$H = \frac{1}{2}gt^2$

$H = \frac{1}{2}g(4.41)^2$

H = 95.3 m