question_answer 54)

                  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 2 seconds). 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 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.  

Answer:

                  Let \[{{h}_{1}}=\] maximum height attained by first ball, \[{{h}_{2}}=\]maximum height attained by second ball using \[{{\upsilon }^{2}}-{{u}^{2}}=2\,gh,\] we get                 But \[{{h}_{1}}-{{h}_{2}}=15\] \[\therefore \] \[{{h}_{2}}=20\] and \[{{h}_{2}}=\text{ }5m\]                 Also using \[h=ut+\frac{1}{2}g{{t}^{2}},\] we get                 \[{{h}_{1}}=2u5\times 4=2u20\]  (where \[t=2s\]) ...(i)                 and \[{{h}_{2}}=\,\frac{u}{2}\,(2-{{t}_{1}})\,-\,5{{(2{{t}_{1}})}^{2}}\]      ?. (ii)                     From eqn. (i) \[2u={{h}_{1}}+20=40\]                 \ \[u=20\text{ }m{{s}^{1}}\]                 This is initial velocity of first ball                 \ Initial velocity of second ball \[=10\text{ }m{{s}^{1}}~\]                 From eqn. (ii), \[{{t}_{1}}=1\text{ }s.\]                 Thus, the exact time interval between the throws of two balls \[=1\,s\].