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- anonymous

a man in a balloon rising vertically with an acceleration 4.9ms release a ball 2 seconds after the balloon is let go from the ground the maximum height above the ground reached by the ball is

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- anonymous

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- anonymous

if the ball is released, its initial speed is 0, and since afterwards it goes down due to gravity, the max height of the ball is the height at which is dropped.
so to get the answer you need to calculate the height of the ballon at t=2 seconds.
it moves with constant accelertion so you can use:
\[y(t)=y_0+v_0t+\frac{1}{2}at^2\]
evaluate for t=2, and think what the values for initial height (y0) and initial speed (v0) should be for the balloon

- BAdhi

@Greg_D At the time of the release of the ball dont you think it will have the velocity of the balloon? So at the height that is calculated from your equation \(y(t)\) it will have a velocity upwards. So before starting to fall the ball will move upwards for a while after it is released.
I think that extra distance should be added as well

- anonymous

yes!
you are right @BAdhi !
is not as i said before, we need first to calculate the speed of the balloon when the ball is released, that will be the initial speed of the ball :
\[v_{balloon}(t) = at\]
evaluate that for t=2, and evaluate y(t) for the balloon at t=2 to get the height.
then for the ball we have:
\[y_{ball}(t)=y_0+v_0t-\frac{1}{2}gt^2\]
with y0, the height of the balloon at t=2 and v0 the speed of the balloon at t=2.
thanks for the correction @BAdhi !

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- BAdhi

\(y_{ball}(t)\) can easily be taken by \(v^2= u^2+2as\), where \(s=y_{ball}\) and \(u=v_{ballon}\). since we do not know the time taken for the ball to reach its maximum height

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