Answer:
As the horizontal velocity does not affect the vertical
motion and initial vertical velocity of both the ball is zero, so both the
balls will be reaching the ground simultaneously i.e. \[{{\text{t}}_{\text{1}}}={{\text{t}}_{\text{2}}}=\sqrt{2h/g}\]
Velocity of the ball dropped vertically while reaching the
ground is \[{{\upsilon }_{\text{1}}}=\sqrt{2gh}\]
For the ball projected horizontally, the horizontal
component velocity \[{{\upsilon }_{{{H}_{2}}}}=u\]and vertical component velocity, \[~{{\upsilon
}_{{{v}_{2}}}}=\sqrt{2gh}\].
So, the total velocity of ball, \[{{\upsilon
}_{2}}=\sqrt{\upsilon _{{{H}_{2}}}^{2}+\upsilon
_{{{V}_{2}}}^{2}}=\sqrt{{{u}^{2}}+2gh}\]which is greater than\[{{\upsilon
}_{1}}\].
It means the ball projected horizontally will strike the
ground, with move speed than the ball dropped vertically downwards.
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