Answer to Question #330206 in Mechanics | Relativity for Shorna

Question

A man standing on the roof of a building throws a ball with a velocity 18.0 ms-1 at an angle 50o from the horizontal. The height of the building is 35.0 m. (Given g = -9.81 ms-2) Calculate:

(i) the maximum height of the ball from the ground,

(ii) the magnitude of the velocity of the ball just before it strikes the ground.

Answer 1

1)

$H=h+\frac{v^2\sin^2\alpha}{2g}=44.7~m.$

2)

$v_x'=v\cos\alpha=11.57~\frac ms,$

$v_y'=\sqrt{2gH}=29.6~\frac ms,$

$v'=\sqrt{v_x'^2+v_y'^2}=31.78~\frac ms.$