The ceiling of a long hall is 25 m high.
Question.
The ceiling of a long hall is $25 \mathrm{~m}$ high. What is the maximum horizontal distance that a ball thrown with a speed of $40 \mathrm{~m} \mathrm{~s}^{-1}$ can go without hitting the ceiling of the hall?

solution:

Speed of the ball, $u=40 \mathrm{~m} / \mathrm{s}$

Maximum height, $h=25 \mathrm{~m}$

In projectile motion, the maximum height reached by a body projected at an angle $\theta$, is given by the relation:

$h=\frac{u^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$

$25=\frac{(40)^{2} \sin ^{2} \theta}{2 \times 9.8}$

$\sin ^{2} \theta=0.30625$

$\sin \theta=0.5534$

$\therefore \theta=\sin ^{-1}(0.5534)=33.60^{\circ}$Horizontal range, $R=\frac{u^{2} \sin 2 \theta}{\mathrm{g}}$

$=\frac{(40)^{2} \times \sin 2 \times 33.60}{9.8}$

$=\frac{1600 \times \sin 67.2}{9.8}$

$=\frac{1600 \times 0.922}{9.8}=150.53 \mathrm{~m}$
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