question_answer

A wooden block is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 100 gm is fired from the foot of the cliff upwards with a velocity of \[\text{1}00\text{ m}{{\text{s}}^{-\text{1}}}\]. At what time will the bullet and block meet each other?

Answer:

Let h be the height and t be the time at which wooden block and bullet will meet. For wooden block: \[s=ut+\frac{1}{2}a{{t}^{2}}\Rightarrow 100-h=\frac{1}{2}g{{t}^{2}}\Rightarrow (1)\] For bullet: In the case of bullet, \[h=100\times t\times \frac{1}{2}g{{t}^{2}}\to (2)\] Substituting (2) in (1) we get \[100-\left( 100t-\frac{1}{2}g{{t}^{2}} \right)=\frac{1}{2}g{{t}^{2}}\] \[100-100+\frac{1}{2}g{{t}^{2}}=\frac{1}{2}g{{t}^{2}}\Rightarrow 100=100t\] \[\Rightarrow t=1s.\]