Question

A ball thrown by one player reaches the other in 2 s. The maximum height attained by the ball above the point of projection will be (Take, g = 10 ms^-2 ) [BHU 2012]

Moderate

Solution

Since, the ball reaches from one player to another in 2 s, so the time period of the flight, T = 2 s
$\Rightarrow \frac{2 u sinθ}{g} = 2 s$
Here, u is the initial velocity and $θ$ is the angle of projection.
$\Rightarrow u sinθ = g . . . . . . (i)$
Now, we know that the maximum height of the projection,
$H = \frac{u^{2} \sin^{2} θ}{2 g} or H = \frac{(u \sin θ)^{2}}{2 g}$
On putting the value of u sin $θ$ from Eq. (i), we get
$H = \frac{g^{2}}{2 g} = \frac{g}{2} or H = \frac{g}{2} = \frac{10}{2} m or H = 5 m$

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