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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)

Arun
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)

 

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

3 years ago
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