Ask a Teacher

Search



ball is thrown vertically upwards with a velocity of 20m/s from the top of high building. The height of point of projection from the ground is 25m. Calculate the maximum height upto which ball rises and the time taken by the ball to hit the ground. (Take g=10m/s square)
  
At the highest point , v = 0
Suppose the ball rises to the height h from the point of projection.
As v2 – u2 = 2gs
Therefore, 02 – 202 = 2 × (- 10) × h
                          or    h = + 20 m


(ii) Net displacement , s = - 25 m

Negative sign is taken because displacement is in opposite direction of initial velocity.

As,  s = ut + ½ gt2

Therefore, - 25 = 20 t + ½ × (-10)t2
or 5 t2 – 20 t – 25 = 0
or         t2 – 4t – 5 = 0
or     (t + 1) (t – 5) = 0

As t ? - 1 , so t = 5 sec


comments powered by Disqus