A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms-1. Then the distance from the top of the tower, at which the two balls meet is

Question

A ball is dropped from the top of a tower of height 100 m and at the same time another ball is projected vertically upwards from ground with a velocity 25 ms-1. Then the distance from the top of the tower, at which the two balls meet is (A) 68.4 m (B) 48.4 m (C) 18.4 m (D) 28.4 m

Tardigrade Admin · Accepted Answer

t=( text height / text speed )= (100/25)=4 s Fall, h =u t+(1/2) g t2 h =0+(1/2) × 9.8 × 42 h =8 × 9.8=78.4 m

Q. A ball is dropped from the top of a tower of height $100 m$ and at the same time another ball is projected vertically upwards from ground with a velocity $25 m s^{- 1}$ . Then the distance from the top of the tower, at which the two balls meet is

68.4 m

48.4 m

18.4 m

28.4 m

78.4 m

$t = \frac{height}{speed} = \frac{100}{25} = 4 s$
Fall, $h = u t + \frac{1}{2} g t^{2}$
$h = 0 + \frac{1}{2} \times 9.8 \times 4^{2}$
$h = 8 \times 9.8 = 78.4 m$

Q. A ball is dropped from the top of a tower of height 100m and at the same time another ball is projected vertically upwards from ground with a velocity 25ms−1. Then the distance from the top of the tower, at which the two balls meet is