# Solve the following :

Question:

Solve the previous problem if the coefficient of restitution is e. Use $\theta=45^{\circ}, e=\frac{9}{4}$ and $h=5 \mathrm{~m}$.

Solution:

$\theta=45, e=\frac{3}{4}$ and $h=5 \mathrm{~m}$

$v=\sqrt{2 g h}$

$=\sqrt{20 \times 5}$

$=10 \mathrm{~m} / \mathrm{s}$

$V_{y}=e v \sin 45^{\circ}$

$V_{y}=\frac{3}{4} \times 10 \times \frac{1}{\sqrt{2}}=\frac{15}{2 \sqrt{2}}$

$V_{x}=v \cos 45^{\circ}$

$=\frac{10}{\sqrt{2}}$

$V=\sqrt{V_{x}^{2}+V_{y}^{2}}=\sqrt{\left(\frac{15}{2 \sqrt{2}}\right)^{2}+\left(\frac{10}{\sqrt{2}}\right)^{2}}=\sqrt{\frac{225+400}{8}}$

$=8.8 \mathrm{~m} / \mathrm{s}$

$=\frac{\pi}{2}-(\theta+\beta)=8^{\circ}$

$x=l \cos \theta \quad y=-l \sin \theta \quad \alpha=-8$

$y=x \tan \alpha-\frac{g x^{2} \sec ^{2} \alpha}{2 u^{2}}$

$-l \sin 45^{\circ}=l \cos 45^{\circ} \tan 8^{\circ}-\frac{g l^{2} \cos ^{2} 45^{\circ} \sec ^{2} 8^{\circ}}{2 \times 8.8}$

$\mathrm{l}=18.5 \mathrm{~m}$