Kinematics 5 Question 11
1. Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$, but different maximum heights $h _1$ and $h _2$. Which of the following is correct?
(12 April 2019 Shift II)
(a) $R^{2}=4 h _1 h _2$
(b) $R^{2}=16 h _1 h _2$
(c) $R^{2}=2 h _1 h _2$
(d) $R^{2}=h _1 h _2$
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Answer:
Correct Answer: 1. (b)
Solution:
Key Idea
Same range for two particles thrown with some initial speed may occurs when they are projected at complementary angle.
$$ \therefore \quad \theta_1+\theta_2=90^{\circ} $$
where, $\theta_1$ and $\theta_2$ are angles of projection.
As maximum range occurs at $\theta=45^{\circ}$ for a given initial projection speed, we take angles of projection of two particles as
$$ \theta_1=45^{\circ}+\theta, \theta_2=45^{\circ}-\theta $$ where, $\theta$ is angle of projectiles with $45^{\circ}$ line. So, range of projectiles will be $$ \begin{aligned} & R=R_1=R_2=\frac{u^2 \sin 2\left(\theta_1\right)}{g} \ & \Rightarrow R=\frac{u^2 \sin 2\left(45^{\circ}+\theta\right)}{g} \Rightarrow R=\frac{u^2 \sin \left(90^{\circ}+2 \theta\right)}{g} \ & \Rightarrow R=\frac{u^2 \cos 2 \theta}{g} \quad \Rightarrow R^2=\frac{u^4 \cos ^2 2 \theta}{g^2} \cdots(i) \end{aligned} $$
Maximum heights achieved in two cases are $$ h_1=\frac{u^2 \sin ^2\left(45^{\circ}+\theta\right)}{2 g} $$ and $$ h_2=\frac{u^2 \sin ^2\left(45^{\circ}-\theta\right)}{2 g} $$
So, $$ h_1 h_2=\frac{u^4 \sin ^2\left(45^{\circ}+\theta\right) \sin ^2\left(45^{\circ}-\theta\right)}{4 g^2} $$
Using $2 \sin A \cdot \sin B=\cos (A-B)-\cos (A+B)$, we have $$ \begin{aligned} \sin \left(45^{\circ}+\theta\right) \sin \left(45^{\circ}-\theta\right) & =\frac{1}{2}\left(\cos 2 \theta-\cos 90^{\circ}\right) \ \Rightarrow \sin \left(45^{\circ}+\theta\right) \sin \left(45^{\circ}-\theta\right) & =\frac{\cos 2 \theta}{2} \quad\left[\because \cos 90^{\circ}=0\right] \end{aligned} $$
So, we have $$ h_1 h_2=\frac{u^4\left(\frac{\cos 2 \theta}{2}\right)^2}{4 g^2} \Rightarrow h_1 h_2=\frac{u^4 \cos ^2 2 \theta}{16 g^2} \cdots(ii) $$
From Eqs. (i) and (ii), we get $$ \Rightarrow \quad h_1 h_2=\frac{R^2}{16} \Rightarrow R^2=16 h_1 h_2 $$
