A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height of 60 m at the time of dropping the ball, how long will the ball take in reaching the ground and what will be its displacement ?

1 Answer
Feb 10, 2018

4.24 s4.24s

Displacement will be 60 m60m

Explanation:

When the ball will be dropped,it will have an initial velocity of 7 m/s7ms upwards,

Now,let it will take time tt to reach the ground,

so using s=ut + 1/2 g t^2s=ut+12gt2

Here, s= 60 ms=60m (as after going up,it will come down by that distance and will go down by the total displacement of 60 m60m) and ,u=-7 m/s,u=7ms (taking upward direction to be positive)

So,our equation becomes, 60= -7t + 1/2 *10 t^260=7t+1210t2

Solving,we get, t=4.24 st=4.24s

ALTERNATIVELY

So,let's calculate the time taken by it to reach the maximum height above the point of release.

using, v=u-g tv=ugt

At its highest point, v=0v=0

So, t= 7/10 s or, 0.7 st=710sor,0.7s

So,after this it will start falling downwards due to the effect of gravity,so net displacement will be 60 m60m ,as it will come down by that distance through which it went upwards after releasing it.

Now to come down by that distance it will take the same amount of time,so to reach its point of release from its highest point,it will take a total of 2*0.7 =1.4 s20.7=1.4s

Let,it will take further time tt to reach the ground after returning to its point of release,

So,using s=ut + 1/2 g t^2s=ut+12gt2

Here, s=60m ,u=7 m/ss=60m,u=7ms (when returning to the point of release,it will gain the same velocity but direction will be reversed.)

So,our equation becomes, 60 = 7t + 1/2 10 t^260=7t+1210t2

Solving,we get, t=2.834 st=2.834s

So,total time required to come back to the ground is 1.4+2.834=4.24 s1.4+2.834=4.24s