First slide

Rectilinear Motion

Question

From the top of the tower of height 400 m, & ball is dropped by a man, simultaneously from the base of the tower, another ball is thrown up with a velocity 50 m/s; at what
distance will they meet from the base of the tower?

Moderate

A

100 m
B

320 m
C

80 m
D

240 m

Solution

Let the first ball meet at a height s from ground
$400 - s = \frac{1}{2} g t^{2}$ ……..(i)
$and for second ball: s = 50 t - \frac{1}{2} g t^{2}$ ……..(ii)
$Adding 50 t = 400, we get$
$t = 8 \sec$
$Now substituting the value of ' t' in (ii),$
$we get s = 50 \times 8 - \frac{1}{2} \times 10 \times 64 = 80 m$

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