Suppose a player hits several baseballs. Whic | vedclass

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(B) The time of flight $T$ for a projectile is given by the formula $T = \frac{2 u \sin 	heta}{g}$.We also know that the maximum height $H$ reached b

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The one which reaches maximum height.

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(B) The time of flight $T$ for a projectile is given by the formula $T = \frac{2 u \sin 	heta}{g}$.We also know that the maximum height $H$ reached by a projectile is $H = \frac{u^2 \sin^2 	heta}{2g}$.From the expression for maximum height,we can write $\sqrt{H} = \frac{u \sin 	heta}{\sqrt{2g}}$,which implies $u \sin 	heta = \sqrt{2gH}$.Substituting this into the time of flight formula: $T = \frac{2 \sqrt{2gH}}{g} = 2 \sqrt{\frac{2H}{g}}$.Since $g$ is constant,$T \propto \sqrt{H}$.Therefore,the baseball that reaches the maximum height will remain in the air for the longest time.

Suppose a player hits several baseballs. Which baseball will be in the air for the longest time?

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