Krishna
0

Given that

Height of the cliff h = 490 m.

Initial speed (Horizontal component) u = v_x = 15 m/s   

Acceleration due to gravity a = g = 9.8 m/s^2

Suitable figure


Step 1: Determine the amount of time the stone takes to hit the ground.

Initial velocity at t =0, v_0 = 0   

Displacement S = h = 490 m

Recall the equations of motion in the plane.

Displacement S = v_0t + \frac{1}{2} gt^2   

490 = 0 * t + \frac{1}{2} (9.8)t^2

t^2 = \frac{490 * 2}{9.8}

t = \sqrt{100}

t = 10 sec

Step 2: Find the final speed of the stone at the ground

Recall the equations of motion in the plane.

Speed along the y-direction v_y^2 = v_0^2 + 2gS

v_y^2 = 0 + 2*9.8*490   

v_y = \sqrt{9604}   

v_y = 98

Thus, Vertical component of the speed v_y = 98 m/s

Magnitude of the final velocity v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{15^2 + 98^2}   

v = 99.14 m/s

Hence, final speed of the stone at the ground v = 99.14 m/s