Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of vi = 10.0 m/s. The cliff is h = 41.9 m above a body of water as shown in the figure below. With what speed and angle of impact does the stone land?

Answer 1

Solution: Initial velocity of the stone v_i = 10.0m/s

Height of the cliff h = 41.9 m

The motion of the stone can be studied in two perpendicular directions; horizontal motion and vertical motion

The initial angle of launch θ = 0^o (as it is thrown horizontally)

Thus the horizontal component of initial velocity v_ix = v_i*cosθ = (10m/s)*cos0^o = 10 m/s

The vertical component of initial velocity v_iy = v_i*sinθ = (10m/s)*sin0^o = 0m/s

Since there is no acceleration in the horizontal direction, the horizontal speed is unchanged till it hits the water surface that is v_ix =v_fx = 10m/s

The gravitational force acts on the stone vertically downwards, thus the vertical displacement of the stone is (from the edge of the cliff to body of water)

y = - 41.9m

let time taken by the stone to fall the height h be t, then

y = v_iy*t + (1/2)*g*t²                                                                                        

-41.9 m = (0m/s)*t + (1/2)*(-9.81m/s²)*t²         (g =-9.81 m/s² as it is acting vertically downward direction)

t² = 2*41.9/9.81

t² = 8.5423

t = 2.9227 s

After t = 2.9227 s the stone strikes the water surface.

Its vertical velocity at the time of impact is given by

v_fy = v_iy + at

v_fy =0m/s + (-9.81m/s²)*2.9227s

v_fy = -28.6717 m/s

The magnitude of the speed of the stone is given by

v = √( v_fx²+ v_fy²)

v = √( 10²+ (-28.6717)²)

v = 30.3655 m/s

And the angle with the horizontal is given by

tanφ = v_fy/v_fx

φ = tan^-1(v_fy/v_fx)

φ = tan^-1 (-28.6717/10)

φ = -70.77^o

This angle is in the fourth quadrant ( -70.77^o eqauals to 289.23^o)

Thus the speed of stone is 30.37 m/s and the angle of impact is -70.77^o