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PART - 1 CH - 3 Motion in a Plane — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 3 Motion in a Plane2 Marks
Question
Two bodies are thrown from the same point with the same velocity at two different angles. Range is same as $R$ for both the bodies and time of flight is $t$ and $t^{\prime}$ then prove that
$R=\frac{1}{2} g t t^{\prime}$
✓

Answer

We know that if any body is thrown at angle $\theta$ and $\left(\frac{\pi}{2}-\theta\right)$ then the range will be equal
$\begin{aligned}
\text { Time of flight } t & =\frac{2 u \sin \theta}{g} \ \ldots(1) \\
\text { Time of flight } t^{\prime} & =\frac{2 u \sin \left(\frac{\pi}{2}-\theta\right)}{g} \\
t^{\prime} & =\frac{2 u \cos \theta}{g} \ \ldots(2)
\end{aligned}$
From equations (1) and (2)
$\begin{aligned}
t t^{\prime} & =\frac{4 u^2 \sin \theta \cos \theta}{g^2} \\
& =\frac{2 u^2 \sin 2 \theta}{g^2} \\
\frac{1}{2} g t t^{\prime} & =\frac{u^2 \sin 2 \theta}{g} \\
R & =\frac{1}{2} g t t^{\prime} \quad \text { Hence Proved. }
\end{aligned}$

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