Question

A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect air resistance.

With what speed was the rock thrown?
How high did the rock travel in the reference frame of the thrower?

Answer 1

Explanation:

Given that,

Distance, d = -56.3 m

It strikes the ground 4.00 seconds after being thrown.

Using second equation of motion to find the speed was the rock thrown. So,

$d=ut+\dfrac{1}{2}at^2$

Here, a = -g

$d=ut-\dfrac{1}{2}gt^2\\\\-56.3=u(4)-4.9(4)^2\\\\-56.3=4u-78.4\\\\u=5.52\ m/s$

Let it will cover a distance of s meters. So,

$s=\dfrac{v^2-u^2}{-2g}\\\\s=\dfrac{0^2-(5.52)^2}{-2\times 9.8}\\\\s=1.55\ m$