Question

A projectile is fired horizontally with a velocity of 98 m/s from the top of a hill 490 m high. Find (1) the velocity with which the projectile strikes the ground. (2) the time taken to reach the ground.

Answer 1

From equations of motion

Taking horizontal component,intial velocity in horizontal direction is 98 while in vertical its 0

$⟹v=\sqrt{u^2+2gh}=\sqrt{{98}^2+2\times 9.8\times 490}$

$⟹v=98\sqrt{2}m/s$

$y=\frac{1}{2}g{t}^2$

$⟹t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 490}{9.8}}$

$⟹t=10s$