A man is a balloon rising vertically with an acceleration of 4.9 m/s square ,releases a ball 2 second after the balloon had been let go from ground . the greatest height above the ground reached by the ball will be
(a)9.3
(b) 19.6
(c) 24.5
(d)14.7
Answers
Let Ub and Vb be the velocities at t=0 and t=2 for the balloon
And accelaration of balloon(Ab)=4.9=g/2 m/s2
Vb=Ub+(Ab)t (Ub=0)
from this we get, Vb=g m/s (Upwards)
and from ,s=(Ub)t+(1/2)(Ab)t^2 (where s is the hieght at which the ball is dropped)
we get s=g meters
when the ball is dropped its initial velocity is gm/s upwards(due to balloon)
hence velocity of ball at ground(using third eq of motion):
V^2=g^2 + 2(g)(g) (since,,initial vel of ball=g m/s,,,acc due to gravity=g m/s2,,,displacement=g meters)
v^2=3g^2------------------------------------1
If we want to find the greatest hieght(to which it rises)then it should rebound back with same velocity(elastic collision)
therfore ,
v^2=u^2-2(g)(H)---------------------------2 (Where H is the max hieght it reaches)
in this case v=0 since at the greatest hieght the vel of ball is zero
and u^2=3g^2 (from eq--1)
therefore
eq 2 becomes:
3g^2=2gs
ie,s=3/2(g)=14.7m
Answer:
(d)14.7
Explanation:
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 .