Answer to Question #272290 in Mechanics | Relativity for Pawi

Question

A stone is thrown vertically up with an initial velocity of 4.9 m/s from the top of a building that is 52 m high. On its way down , it misses the top of the building and goes straight to the ground. Find a.) maximum height reached b.) maximum height reached relative to the ground and its velocity when it strikes the ground if the time of flight is 3.8 s. 

Answer 1

a) $h=\frac{v_0^2}{2g}=\frac{4.9^2}{2\cdot9.8}=1.225\ (m)$ maximum height reached relative to the top of the building.

b) $H=52+1.225=53.225\ (m)$ maximum height reached relative to the ground.

$H=\frac{v^2}{2g}\to v=\sqrt{2gH}=\sqrt{2\cdot9.8\cdot53.225}=32.3\ (m/s)$