A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff in the upward direction with a velocity of $100 m s^{- 1}$ . At what time will the bullet and the block meet ?

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Solution

Let $s_{1}$ and $s_{2}$ be the displacements covered by wooden block and bullet respectively.
Given. height of the tower is 100 m
Therefore, $s = s_{1} + s_{2} = 100 m$ .
From the second equation of motion,
$s = u t + \frac{1}{2} \times {a t}^{2}$
Before the collision, in the case of block, initial velocity, u = 0 and the acceleration, a = g (acceleration due to gravity)
Therefore, $s_{1} = \frac{1}{2} g t^{2} = \frac{1}{2} \times 9.8 \times t^{2} = 4.9 t^{2}$
In the case of bullet, initial velocity, $u = 100 m s^{- 1}$ and the acceleration, $a = - g$ (acceleration due to gravity)
$s_{2} = u t - \frac{1}{2} g t^{2}$
$= 100 t - \frac{1}{2} \times 9.8 \times t^{2} = 100 t - 4.9 t^{2}$
Substituting the values in equation, $s_{1} + s_{2} = 100 m$ , we get,
$4.9 t^{2} + 100 t - 4.9 t^{2} = 100$
or, 100t = 100
or, t = 1 s

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