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A particle is projected from the ground with some initial velocity making an angle of $45^{\circ}$ with the horizontal. If it reaches a height of $7.5 \mathrm{~m}$ above the ground, while it travels a horizontal distance of $10 \mathrm{~m}$ from the point of projection, then the initial speed of particle is (assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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The correct answer is:
$20 \mathrm{~m} / \mathrm{s}$
The given situation is as shown in figure,
Let the particle reaches point $P$ in time $t$. Then, horizontal distance,
$x=u \cos \theta t \Rightarrow 10=u \cos 45^{\circ} \times t$
$\Rightarrow \quad t=\frac{10 \sqrt{2}}{u}$
and vertical distance,
$\begin{aligned} y & =u \sin \theta t-\frac{1}{2} g t^2 \\ \Rightarrow \quad 7.5 & =u \sin 45^{\circ} \times t-\frac{1}{2} \times 10 \times t^2\end{aligned}$
Substituting value of $t$ from Eq. (i) into Eq. (ii), we get
$\begin{array}{ll}\Rightarrow & 7.5=\frac{u}{\sqrt{2}} \times \frac{10 \sqrt{2}}{u}-\frac{1}{2} \times 10 \times\left(\frac{10 \sqrt{2}}{u}\right)^2 \\ \Rightarrow & 7.5=10-\frac{1000}{u^2} \\ \Rightarrow & u^2=\frac{1000}{2.5}=400 \text { or } u=20 \mathrm{~m} / \mathrm{s}\end{array}$
Let the particle reaches point $P$ in time $t$. Then, horizontal distance,
$x=u \cos \theta t \Rightarrow 10=u \cos 45^{\circ} \times t$
$\Rightarrow \quad t=\frac{10 \sqrt{2}}{u}$
and vertical distance,
$\begin{aligned} y & =u \sin \theta t-\frac{1}{2} g t^2 \\ \Rightarrow \quad 7.5 & =u \sin 45^{\circ} \times t-\frac{1}{2} \times 10 \times t^2\end{aligned}$
Substituting value of $t$ from Eq. (i) into Eq. (ii), we get
$\begin{array}{ll}\Rightarrow & 7.5=\frac{u}{\sqrt{2}} \times \frac{10 \sqrt{2}}{u}-\frac{1}{2} \times 10 \times\left(\frac{10 \sqrt{2}}{u}\right)^2 \\ \Rightarrow & 7.5=10-\frac{1000}{u^2} \\ \Rightarrow & u^2=\frac{1000}{2.5}=400 \text { or } u=20 \mathrm{~m} / \mathrm{s}\end{array}$
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