a man in a balloon rising vertically with an acceleration of 4.9 metre per second releases a ball 2 second after the balloon is released from the ground time for the released ball to reach the ground is(a)2.7s(b)3.4s(c)1s(d)2.3s
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Explanation:
(diagram is attached)
after 2seconds velocity of man
v= u+at
= 0+ 4.9×2
= 9.8 m/second in upward direction
height H1 =( 1/2)×4.9×4= 9.8 m
time taken to reach max height
=> 0= - 9.8+9.8×t1
=> t1= 1 sec.
height H2=>. v²= u²-2gh=> 0=( 9.8)²-2×9.8×H2
=> H2= 4.9m
total height( H )= 9.8+4.9= 14.7 m
14.7= (1/2) ×9.8 t² = 4.9 t²
=> t²= 14.7/4.9= 3
=> t= √3 seconds
total time = t1+t= 1+√3 sec
= 1+1.732= 2.732 sec
=> 2.7 sec option (a) is correct
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