Question
Solve the previous problem if the coefficient of restitution is e. Use $\theta=45^\circ,\ \text{e}=\frac{3}{4}$ and h = 5m
✓
Answer
$\text{h}=5\text{m},\ \theta=45^\circ,\ \text{e}=\Big(\frac{3}4{}\Big)$Here the velocity with which it would strike $=\text{v}=\sqrt{\text{2g}\times5}=10\text{m/sec}$
After collision, let it make an angle $\beta$ with horizontal. The horizontal component of velocity $10\cos45^\circ$ will remain unchanged and the velocity in the perpendicular direction to the plane after wllisine.
$\Rightarrow\text{V}_{\text{y}}=\text{e}\times10\sin45^\circ$
$=\Big(\frac{3}{4}\Big)\times10\times\frac{1}{\sqrt2}=(3.75)\sqrt2\text{m/sec}$
$\Rightarrow\text{V}_{\text{x}}=10\cos45^\circ=5\sqrt2\text{m/sec}$
So, $\text{u}=\sqrt{\text{V}_{\text{x}}^2+\text{V}_{\text{y}}^2}$
$=\sqrt{50+28.125}=\sqrt{78.125}=8.83\text{m/sec}$
Angle of reflection from the wall $\beta=\tan^{-1}\Big(\frac{3.75\sqrt2}{5\sqrt2}\Big)=\tan^{-1}\Big(\frac{3}{4}\Big)=37^\circ$
⇒ Angle of projection $\alpha=90-(\theta+\beta)=90-(45^\circ+37^\circ)=8^\circ$
Let the distance where it falls = L
$\Rightarrow\text{x}=\text{L}\cos\theta,\ \text{y}=-\text{L}\sin\theta$
Angle of projection $(\alpha)=-8^\circ$
Using equation of trajectory, $\text{y}=\text{x}\tan\alpha-\frac{\text{gx}^2\sec^2\alpha}{2\text{u}^2}$
$\Rightarrow-\ell\sin\theta=\ell\cos\theta-\tan8^\circ-\frac{\text{g}}{2}\times\frac{\ell\cos^2\theta\sec^28^\circ}{\text{u}^2}$
$\Rightarrow-\sin45^\circ=\cos45^\circ-\tan8^\circ-\frac{10\cos^245^\circ\sec8^\circ}{(8.83)^2}(\ell)$
Solving the above equation we get,
$\ell=18.5\text{m}$
After collision, let it make an angle $\beta$ with horizontal. The horizontal component of velocity $10\cos45^\circ$ will remain unchanged and the velocity in the perpendicular direction to the plane after wllisine.
$\Rightarrow\text{V}_{\text{y}}=\text{e}\times10\sin45^\circ$
$=\Big(\frac{3}{4}\Big)\times10\times\frac{1}{\sqrt2}=(3.75)\sqrt2\text{m/sec}$
$\Rightarrow\text{V}_{\text{x}}=10\cos45^\circ=5\sqrt2\text{m/sec}$
So, $\text{u}=\sqrt{\text{V}_{\text{x}}^2+\text{V}_{\text{y}}^2}$
$=\sqrt{50+28.125}=\sqrt{78.125}=8.83\text{m/sec}$
Angle of reflection from the wall $\beta=\tan^{-1}\Big(\frac{3.75\sqrt2}{5\sqrt2}\Big)=\tan^{-1}\Big(\frac{3}{4}\Big)=37^\circ$
⇒ Angle of projection $\alpha=90-(\theta+\beta)=90-(45^\circ+37^\circ)=8^\circ$
Let the distance where it falls = L
$\Rightarrow\text{x}=\text{L}\cos\theta,\ \text{y}=-\text{L}\sin\theta$
Angle of projection $(\alpha)=-8^\circ$
Using equation of trajectory, $\text{y}=\text{x}\tan\alpha-\frac{\text{gx}^2\sec^2\alpha}{2\text{u}^2}$
$\Rightarrow-\ell\sin\theta=\ell\cos\theta-\tan8^\circ-\frac{\text{g}}{2}\times\frac{\ell\cos^2\theta\sec^28^\circ}{\text{u}^2}$
$\Rightarrow-\sin45^\circ=\cos45^\circ-\tan8^\circ-\frac{10\cos^245^\circ\sec8^\circ}{(8.83)^2}(\ell)$
Solving the above equation we get,
$\ell=18.5\text{m}$
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