Question

The ceiling of a long hall is $25\text{m}$ high. What is the maximum horizontal distance (in $textm$) that a ball thrown with a speed of $40\text{m/s}$ from the floor can go without hitting the ceiling of the hall?
(Take $g=10{\text{m/s}}^2$ and answer upto 2 decimal places)

Answer 1

Height of the ceiling, $H=25\text{m}$
Initial speed of the ball, $u=40\text{m/s}$.
We know that maximum height obtained by a projectile is given by:
$h_0=\frac{u^2{\sin }^2\theta }{2g}$

If the projectile just grazes the ceiling (without hitting it),
$25=\frac{(40{)}^2{\sin }^2\theta }{2\times 10}$
or ${\sin }^2\theta =\frac{25\times 2\times 10}{40\times 40}=\frac{5}{16}$
$\Rightarrow \sin \theta =\sqrt{\frac{5}{16}}=\frac{\sqrt{5}}{4}$ and $\cos \theta =\frac{\sqrt{11}}{4}$

Then, horizontal range of the ball is given by
$R=\frac{u^2\times \sin 2\theta }{g}=\frac{u^2\times 2\sin \theta \cos \theta }{g}$
$=\frac{1600\times 2\times \sqrt{5}\times \sqrt{11}}{16\times 10}=\frac{1483.23}{10}=148.32\text{m}$