Q. A balloon is moving up with a speed 60 m/s.When the balloon is at a height 225 m above the ground a stone is thrown horizontally with a speed 40 m/s with respect to balloon.

Find:-

a) The horizontal displacement of the stone when it is at the top of its trajectory.

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At the moment of throwing, the stone will possess two components of velocity,

1. Upward, 60m/s

2. Horizontal, 40 m/s

At the top of its trajectory, the vertical component of velocity is zero. Applying laws of motion for this component,

0=60-gt

t = 60/g =60/9.81=6.1155 s

This is the time to reach the top of its trajectory.

a) Horizontal range covered by the stone in this time = horizontal velocity * time

=40*6.1155

=244.6 m.

Therefore, the horizontal displacement of the stone when it is at the top of its trajectory is 244.6 m.

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