Question

A projectile is launched horizontally from the top of a vertical cliff with a velocity of 15m/s is observed to strike the ground at a distance of 45m from the foot of the cliff. What is the angle the path of the projectile makes with the ground after the moment of impact

Answer 1

Range =45m

velocity

v_h=15m/sec

Newton's motion 1st law

$v=u+at$

Put u=0,a=g

$v_v=0+9.8\times3=29.43m/sec$

Horizontal velocity

v_h=15m/sec

Angle

$$\theta=tan^{-1}(\frac{v_v}{v_h})=tan^{-1}(\frac{29.43}{15})=62.99°$$