Question

A man stands at the edge of a cliff and throws a rock horizontally over the edge with a speed of

v₀ = 15.5 m/s.

The rock leaves his hand at a height of

h = 46.0 m

above level ground at the bottom of the cliff, as shown in the figure. Note the coordinate system in the figure, where the origin is at the bottom of the cliff, directly below where the rock leaves the hand

m

h too 0

How long (in s) after being released does the rock strike the ground below the cliff?

With what speed (in m/s) and angle of impact (in degrees clockwise from the +x-axis) does the rock land?

Answer 1

For time, apply equation of motion in y direction,

h = gt²/2

Therefore, t = sqrt(2h/g) ≈ 3sec

For final speed, apply work energy theorem,

mgh = mv²/2 - mv₀²/2

On solving, we get

v = 34 m/s

And angle with +x axis in clockwise direction is

Angle = cos¯¹(v₀/v) ≈ 63°