Question

A stone is thrown from the top of a building 30m high. If it strikes the ground at an angle 45, with what speed was it thrown?

Answer 1

Answer:

24.2 m/s

Explanation:

The stone strikes the ground at an angle of 45 degrees: this means that its vertical velocity is equal (in magnitude) to its horizontal velocity, in fact:

$tan \theta = \frac{|v_y|}{v_x}\\tan 45^{\circ} = 1 \rightarrow |v_y| = v_x$

The motion along the vertical direction is a uniformly accelerated motion, so we can find the final vertical velocity using the following suvat equation

$v_y^2 -u_y^2 = 2as$

where

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the stone is thrown horizontally)

$a=g=9.8 m/s^2$ is the acceleration of gravity (we take downward as positive direction)

s = 30 m is the vertical displacement

Solving for vy,

$v_y = \sqrt{u_y^2+2as}=\sqrt{0+2(9.8)(30)}=24.2 m/s$

This means that the horizontal velocity is also 24.2 m/s: and since the horizontal velocity is constant during the whole motion (there is no acceleration in the horizontal direction), this means that the stone was thrown exactly at 24.2 m/s.