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?
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?
Updated On: Jun 23, 2024
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Solution and Explanation
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
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
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