Answer to Question #163196 in Mechanics | Relativity for Olawale Quadri

Question

A stone of mass 1kg is thrown at 10m/s upward Making an angle 37° with the horizontal from a building thta is 20m high.using law of conservation of energy . calculate the speed when the stone hits the ground.

Answer 1

By the law of energy conservation :

$\frac{mv^2}{2}+mgh=\frac{mu^2}{2}$

where, $v$ is inutial speed, $u$ is final speed, $h$ is building height

When the stone falls back down and reaches the horizontal position again, it will have the same angle as at launch: $37\degree$ .

Horizontal component of velocity:

$v_x=u_x=10cos37\degree=8$ m/s; this does not change

Vertical component of initial velocity:

$v_y=10sin37\degree=6$ m/s

So:

$u_y=\sqrt{v^2_y+2gh}=\sqrt{6^2+2\cdot9.8\cdot20}=20.7$ m/s

The final speed:

$u=\sqrt{8^2+20.7^2}=22.2$ m/s