question_answer

A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Find when and where the two stones meet \[(\text{g}=\text{9}.\text{8 m}{{\text{s}}^{\text{-2}}}).\]

Answer:

                Height of the tower = 100 m (i) For the falling ball: Height, \[g=100\,\,m\] Initial velocity, \[u=0\,\,m/s\] Acceleration, \[a=g\,\,m/{{s}^{2}}\] (ii) For the ball thrown upwards: Initial velocity, \[u=25\,\,m/s\] Acceleration, \[a=-g\,\,m/{{s}^{2}}\] Let the two balls meet at time t seconds after the start and at a distance x m from the ground and\[(100-x)m\]from the top of the tower. For the ball falling downwards: From the equation, \[h=ut+{\scriptstyle{}^{1}/{}_{2}}\,\,g{{t}^{2}}\] Substituting the values, we get, \[(100-x)=ut+\frac{1}{2}g{{t}^{2}}\] \[=(0)t+\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}^{2}}\] For the ball thrown upwards: From the equation, \[h=ut-{\scriptstyle{}^{1}/{}_{2}}g{{t}^{2}}\] Substituting the values, we get, \[2(u+g)=u+\frac{5\times g}{2}\] Adding equations (1) and (2), \[x=25\times t\,\,or\,\,t=4s\] Thus, the two balls meet 4 seconds after the start. Substituting the value of t in Eq. (2), we get \[x=25m/s\times 4s\frac{1}{2}\times 9.8m/{{s}^{2}}\times {{4}^{2}}{{s}^{2}}\] \[=100m-78.4=21.6m\] Thus, the two balls meet 4 seconds after the start at a distance of 21.6 m from the ground