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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
$$
\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)
$$
Options:
$$
\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)
$$
Solution:
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Verified Answer
The correct answer is:
$5 \mathrm{~m}$
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, $\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{(10)^2}{2 \times 10}=5 \mathrm{~m}$
$\therefore \mathrm{u} \sin \theta=\mathrm{g}=10$
Maximum height, $\mathrm{H}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}=\frac{(10)^2}{2 \times 10}=5 \mathrm{~m}$
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