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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» <html><body><p>`2.0 ms^(-1)`<br/>`12 m s^(-1)`<br/>`25 ms^(-1)`<br/>`15 m s^(-1)` </p>Solution :Speed in horizontal direction remains constant during whole <a href="https://interviewquestions.tuteehub.com/tag/journey-1062300" style="font-weight:bold;" target="_blank" title="Click to know more about JOURNEY">JOURNEY</a> because there is no acceleration in this direction. <br/> So, `v_1 = <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> ms^(-1)` <br/> In vertical direction <br/> Loss of <a href="https://interviewquestions.tuteehub.com/tag/gravitational-476409" style="font-weight:bold;" target="_blank" title="Click to know more about GRAVITATIONAL">GRAVITATIONAL</a> potential energy = gain in kinetic energy <br/> `:. mgh = 1/2 mv_2^2` <br/> or `v_2^2 = 2gh = 2 xx 10 xx (70 - 60) = <a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>` <br/> Hence , the speed with which he touches the cliff B is <br/> `v= sqrt(v_1^2 + v_2^2) = sqrt(25 + 200) = sqrt(<a href="https://interviewquestions.tuteehub.com/tag/225-294624" style="font-weight:bold;" target="_blank" title="Click to know more about 225">225</a>) = 15 ms^(-1)`.</body></html>


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