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A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2s). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is 15 m at t = 2. The gap is found to remain constant. The velocities with which the balls were thrown are (Take g = 10 m s^(-1))

Answer» <html><body><p>`20 m s^(-1) , 10 m s^(-1)`<br/>`10 m s^(-1) , 5 m s^(-1)`<br/>`16 m s^(-1) , 8 m s^(-1)`<br/>`30 m s^(-1), 15 m s^(-1)`</p>Solution :For first <a href="https://interviewquestions.tuteehub.com/tag/stone-1228007" style="font-weight:bold;" target="_blank" title="Click to know more about STONE">STONE</a>, <br/> taking the vertical upwards motion of the first stone up to highest point <br/> Here, u = `u_1`, v = 0 (At highest point velocity is zero) <br/> a = -g, S = `h_1` <br/> As `v^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) - u^(2) = 2aS``<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> (0)^(2) - u_(1)^(2) = 2(-g)h_1`or`h_1 = u_(1)^(2)/(2g)`...(i)<br/> For second stone, <br/> Taking the vertical upwards motion of the second stone up to highest point <br/> here, u = `U_2, v = 0, a = -g, S = h_2` <br/> As `v^(2) - u^(2)` = 2as <br/> `therefore(0)^(2) - (u_2)^(2) = 2(-g)h_2`or`h_2 = u_(1)^(2)/(2g)`.............(ii) <br/> As per question <br/> `H_1 - h_2 = 15 m , u_2 = u_1/2` <br/> <a href="https://interviewquestions.tuteehub.com/tag/subtract-1231765" style="font-weight:bold;" target="_blank" title="Click to know more about SUBTRACT">SUBTRACT</a> (ii) from (i), we get, `h_1 - h_2 = u_(1)^(2) /(2g)-(u_(2)^(2)/(2g))`<br/> On substituting the <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> information, we get <br/> `15 = u_(1)^(2)/(2g)-u_(1)^(2)/(2g)=(3u_(1)^(2))/(8g)` or `u_(1)^(2)=(15 xx 8g)/(3)=(15 xx 8 xx 10)/(3)= 400` <br/> or`u_1 = 20 m s^(-1)` and`u_2 = U_1/2 = 10 m s^(-1)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_PHY_XI_C03_E01_065_S01.png" width="80%"/></body></html>


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