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Q...A man in a balloon rising vertically with an acceleration of 4.9 m/s2 releases a ball 2 seconds after the balloon is let go from the ground. The greatest height above the ground reached by the ball is:
a. 14.7 m b. 19.6 m c. 9.8 m d. 24.5 m
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
b is the answer. is am i right
while releasing the ball the balloon was at a height of 9.8m from the ground and was rising up at 9.8m/s.
so the maximum height raised by the ball is 9.8 + (9.8)2/2*9.8 = 14.7m
the answer is a. 14.7m
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