A motor cyclist is trying to jump across a path as shown by driving horizontally off a cliff A at a speed of 5 ms-l. Ignore air resistance and take g = l0 ms-2. Th" speed with which he rouches the peak -B is:

Speed in horizontal direction remains constant during whole journey because there is no acceleration in this direction.So, vh = 6 ms-1In vertical direction:Loss of gravitation potential energy = gain in KEi.e., mgh = 12mvV2        vV2 = 2gh = 2×10×(70-60) = 200Hence, the speed with which he touches the cliff B isv = vB2+vV2 = 25+200 = 225 = 15 ms-1