First slide

Rectilinear Motion

Question

From the top of the tower of height 400 m, a 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: $s = 50 t - \frac{1}{2} g t^{2}$
Adding $50 t = 400$ , we get
t= 8 sec.
Now substituting the value of ‘t’ in (i),
$we get s = 50 \times 8 - \frac{1}{2} \times 10 \times 64 = 80 m$

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