A motor cyclist is trying to jump across a path as shown by driving horizontally off a cliff A at a speed of 5 m s^(-1). Ignore air resistance and take g = 10 m s^(-2). The speed with which he touches the cliff B is

A motor cyclist is trying to jump across a path as shown by driving ho

1.	A motor cyclist is trying to jump across a path as shown by driving horizontally off a cliff A at a speed of 5 m s^(-1). Ignore air resistance and take g = 10 m s^(-2). The speed with which he touches the cliff B is
Answer» `2.0 ms^(-1)` `12 m s^(-1)` `25 ms^(-1)` `15 m s^(-1)` Solution :Speed in horizontal direction remains constant during whole JOURNEY because there is no acceleration in this direction. So, `v_1 = 5 ms^(-1)` In vertical direction Loss of GRAVITATIONAL potential energy = gain in kinetic energy `:. mgh = 1/2 mv_2^2` or `v_2^2 = 2gh = 2 xx 10 xx (70 - 60) = 200` Hence , the speed with which he touches the cliff B is `v= sqrt(v_1^2 + v_2^2) = sqrt(25 + 200) = sqrt(225) = 15 ms^(-1)`.