In the figure a stone is projected at a cliff of height h wi

In the figure, a stone is projected at a cliff of height h with an initial speed of 45.0 m/s directed at an angle theta_0 = 54.0 degree above the horizontal. The stone strikes at A, 5.10 s after launching. Find the height h of the cliff, the speed of the stone just before impact at A, and the maximum height H reached above the ground.

Solution

(a)The height of the cliff h is given by the equation of motion

h= y₀ + u_y0t – 1/2gt²

y₀ = Initial diplacement in y-direction =0

where u_y0= initial velocity of projection in y-direction = u sin 54°

Thus h= 45 sin 54°x5.10 – 1/2x9.8x5.1²= 58.22m

(b)First, we have to determine the velocity components at the moment of the impact.

v_xt = u_x0 = 45 cos 54° =26.45m/s

v_yt = u_y0 – gt = -13.57 m/s

Thus, we get the speed as magnitude of velocity components

V= ( v_xt² + v_yt²) = 29.72 m/s

(c)We have maximum height H= u_y0²/2g = 67.62m