Answer to Question #298505 in Mechanics | Relativity for Sunil

Question

A ball is thrown forward horizontally from the top of a cliff with a velocity 10ms-1. The

height of the cliff above the ground is 45m. Calculate (i) the time to reach the ground (ii)

the distance from the cliff of the ball on hitting the ground (iii) the velocity with which

the ball hits the ground. (iv) the direction of the ball to the horizontal just before it hits

the ground

Answer 1

$$t=\sqrt{2\frac{45}{9.8}}=3.0\ s\\d=(10)(3)=30\ m\\ \theta=\arctan{\frac{9.8\cdot3}{10}}=71\degree\text{ with horizontal}$$