Question

A motorcyclist is trying to jump across a path as shown by driving horizontally off a cliff A at a speed of 5 m/s. Ignore air resistance and take g = 10m/s2 . The speed with which he touches the cliff B is?

Answer 1

Given- vh = 5 ms-1 and g = 10m/s2
According to the law of conservation of energy
Loss of gravitational potential energy = Gain in Kinetic Energy
mgh = $\frac {1}{2}$mvv2
vv2 = $\frac {2mgh}{m}$
vv2 = 2gh
vv2 = 2 x 10 x (70-60)
vv2 = 200 ms-1
speed with which he touches the cliff B = $\sqrt {v_h^2 + v_v^2}$ = $\sqrt {5^2 + 200}$ = $\sqrt {225}$ = 15 ms-1