Kinematics Ques 4
- A shell is fired from a fixed artillery gun with an initial speed $u$ such that it hits the target on the ground at a distance $R$ from it. If $t_1$ and $t_2$ are the values of the time taken by it to hit the target in two possible ways, the product $t_1 t_2$ is
(2019 Main, 12 April I)
(a) $\frac{R}{4 g}$
(b) $\frac{R}{g}$
(c) $\frac{R}{2 g}$
(d) $\frac{2 R}{g}$
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Answer:
Correct Answer: 4.( d )
Solution:
Formula:
Given, range of the fired shell, $ R=R $ and time of flights are $t_1$ and $t_2$.
Let $\theta_1$ and $\theta_2$ are the two angles at which shell is fired. As, range in both cases is same, i.e.
$ \begin{aligned} & R_1=R_2=R \\ & \text { Here, } \\ & R_1=\frac{u^2 \sin 2 \theta_1}{g} \\ & \text { and } \quad R_2=\frac{u^2 \sin 2 \theta_2}{g} \\ & \Rightarrow \quad R=\frac{u^2 \sin 2 \theta_1}{g}=\frac{u^2 \sin 2 \theta_2}{g} \\ & \Rightarrow \quad \sin 2 \theta_1=\sin 2 \theta_2 \\ & \Rightarrow \quad \sin 2 \theta_1=\sin \left(180-2 \theta_2\right)[\because \sin (180-\theta)=\sin \theta] \\ & \Rightarrow \quad 2\left(\theta_1+\theta_2\right)=180 \text { or } \theta_1+\theta_2=90 \\ & \Rightarrow \quad \theta_2=90-\theta_1 \\ & \end{aligned} $
So, time of flight in first case,
$ t_1=\frac{2 u \sin \theta_1}{g} $
and time of flight in second case,
$ t_2=\frac{2 u \sin \theta_2}{g}=\frac{2 u \sin \left(90-\theta_1\right)}{g}=\frac{2 u \cos \theta_1}{g} $
From Eqs. (iii) and (iv), we get
$ \begin{aligned} t_1 t_2 & =\frac{2 u \sin \theta_1}{g} \times \frac{2 u \cos \theta_1}{g} \\ \Rightarrow \quad t_1 t_2 & =\frac{4 u^2 \sin \theta_1 \cos \theta_1}{g^2} \\ \Rightarrow \quad t_1 t_2 & =\frac{2 u^2 \sin 2 \theta_1}{g^2}(\because \sin 2 \theta=2 \sin \theta \cos \theta) . \end{aligned} $
From Eq. (i), we get
$ \begin{aligned} R & =\frac{u^2 \sin 2 \theta_1}{g} \\ \therefore \quad t_1 t_2 & =\frac{2 R}{g} \end{aligned} $
BETA
