A cricket player hit a ball like a projectile, but the fielder caught the ball after 2 second. The maximum…

A cricket player hit a ball like a projectile, but the fielder caught the ball after 2 second. The maximum height reached by the ball is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$
  1. $2 \mathrm{~m}$
  2. $5 \mathrm{~m}$
  3. $4 \mathrm{~m}$
  4. $3 \mathrm{~m}$

Solution

Time of flight, $\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}=2 \mathrm{~s}$ $\therefore \mathrm{u} \sin \theta=\mathrm{g}=10$ Maximum height, $\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{(10)^2}{2 \times 10}=5 \mathrm{~m}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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