A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find

(a) The time it tskes to reach the ground,

(b) The horizontal distance it travels before reaching the ground,

(c) The velocity (direction and magnitude) with which it strikes the ground.

It is case of projective fired horizontally from a height h.

g = 9.8 $m/s^2$

(a) time taken to reach the ground,

t' =$\sqrt{\left(\frac{2h}{g}\right)}$

= $\sqrt{\frac{2\times 100}{9.8}}$ = 4.51 sec

(b) Horizontal range

x= ut = $20\times 4.5$ = 9.0 m

(c) Horizontal velocity remains constant through out the motion.

At 'A', $V_x$ = 20 m/s

$v_y$ = u + at = 0 + 9.8 $\times 4.5$

= 44.1 m/s

Resultant velocity,

$v_r=\sqrt{(44.1{)}^2+{20}^2}$

tan $\beta$ = $\frac{v^y}{v^x}=\frac{44.1}{20}=2.205$

$\Rightarrow \beta =ta{n}^{-1}\left(2.205\right)={66}^0$

The ball strikes the ground with a velocity of 48.42 m/s at an angle $\theta ={66}^0$ with horizontal.