First slide

Projection Under uniform Acceleration

Question

A ball thrown by one player reaches the other in 2 sec. The maximum height attained by the ball above the point of projection will be about

Easy

A

2.5 m
B

5 m
C

7.5 m
D

10 m

Solution

$\begin{array}{l} t = \frac{2 u \sin θ}{g} or 2 = \frac{2 u \sin θ}{g} \\ ∴ u \sin θ = g \\ ∴ h_{m} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{g^{2}}{2 g} = \frac{g}{2} = 5 m . \end{array}$

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