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Grade: 12th Pass


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

3 years ago

Answers : (1)

24742 Points

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)




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)




eq 2 becomes:





3 years ago
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