Question

A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50 meter/sec. from the surface of the tower, then they will meet at which height from the surface of the tower

• A. 100 meters
• B. 320 meters
• C. 80 meters
• D. 240 meters

Answer 1

Answer: (C)

80 meters

Answer 2

Let both balls meet at point P after time t.

The distance travelled by ball A ($h_{1}) = \frac{1}{2}gt^{2}$ .....(i)

The distance travelled by ball B $(h_{2}) = ut - \frac{1}{2}gt^{2}$.....(ii)

By adding (i) and (ii) $h_{1} + h_{2} = ut$ = 400

(Given $h = h_{1} + h_{2} = 400.$)

$\therefore t = 400/50 = 8sec$ and $h_{1} = 320m\text{, }h_{2} = 80m$