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Grade 11Mechanics

Two hot air balloons A and B move in a vertical upward direction with equal uniform velocities of 5m/s along the same line initially ballon A is on the ground and balloon B is at a height of 80m from the ground. A stone is dropped from the ballon B at time t =0 . The time after the stone reaches ballon A is sec . The value of the is

Profile image of Dheeraj
8 Years agoGrade 11
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1 Answer

Profile image of Piyush
8 Years ago
As the stone is released from balloon it have velocity in vertically upward direction.so, stone will be at distance of 80 m from ground after 1 sec and it will have velocity of 5 m/s.
In that 1 sec the balloon will travel 5 m.so separation between them would be 75m.
Let S1 distance is travelled by stone and S2 distance is travelled by balloon.
 
S1 + S2=75
5t +1/2*10*t^2 +5t= 75
After solving the equation 
t=3 sec
I.e the time taken by stone to meet ballon A is 3+1= 4 sec