A man in a balloon rising vertically with an acceleration of 4.9m/s^2release a ball 2second after the balloon is let go from the ground. The greatest height above the ground reached by the ball is (9.8m/s^2)
Answers
Answer:
Assuming that the balloon was at rest on the ground
It's initial vertical velocity ( say u ) = 0
The given acceleration of balloon is supposed to be the NET acceleration of the balloon , and that it is obtained after considering the effect of acceleration due to gravity.
So, net acceleration of the balloon ( say a ) = 4.9 m/s² vertically upwards.
Here t = 2s being measured from the instant the balloon starts to rise from the ground.
Using x = u*t + 0.5 * a * t²
Or
x = ( 0*2 +0.5*4.9*2² ) m
= ( 0 + 9.8 ) m = 9.8 m.
When the ball is released from the balloon 2 s after it starts to rise,
the speed of the ball would be same as that of the balloon and in the same direction too.
Speed of balloon at t = 2s = ( 0 + 4.9*2 ) m/s
= 9.8 m/s .
So speed of ball at the point of release = 9.8 m/s vertically upwards ( say v )
So max height reached by the ball,
measured from the point of RELEASE
=v²/2g = { (9.8)²/(2*9.8) } m = 4.9 m
So max. height of ball as measured from the ground = ( 4.9 + 9.8 ) m = 14.7 m