Question

from the top of the tower of height 400 m a ball is dropped by man simultaneously from the base of the tower another ball is thrown up with velocity 50 M per second at what distance will they meet from the base of the tower

Answer 1

first we will take out relative speed.
the balls are moving opposite to each other
Therefore relative speed =50+0=50m/s.
distance =400m.
time taken by balls to meet together=400÷50=8seconds.
s=ut-1/2at^2 (motion formula against gravity).
u=50m/s.
t=8seconds.
a=10m/s^2.
putting these values to formula.
50×8-1/2×10×(8)^2.
400-64×5.
400-320=80m.
the balls will meet 80m above ground.

Answer 2

Answer:

Required distance is 80 m

Explanation:

Let,

After 't' seconds they meet.

So, Dropping ball 's' after t second =

$ut + \frac{1}{2} g {t}^{2} = \frac{1}{2} \times \times 10 \times {t }^{2} = 5 {t}^{2}$

Uplifting ball 's' after t second =

$400 - 5 {t}^{2}$

Or,

$ut - \frac{1}{2} g {t}^{2} = 50 - 5 {t}^{2}$

Now,

$400 - 5 {t}^{2} = 50t - 5 {t}^{2} \\ t = 8$

Required distance = (400-320)= 80 m