Question

Three stones $A,B,C$ are projected from surface of very long inclined plane with equal speeds and different angles of projection as shown in figure. The incline makes an angle $\theta$ with horizontal. If $H_A,H_B$ and $H_C$ are maximum height attained by $A,B$ and $C$ respectively above inclined plane then: (Neglect air friction)

• A. $H_A+H_C=H_B$
• B. $H_A^2+H_C^2=H_B^2$
• C. $H_A+H_C=2H_B$
• D. $H_A^2+H_C^2=2H_B^2$

Answer 1

The correct option is A $H_A+H_C=H_B$

$\begin{array}{c}{H}_B=\frac{u^2}{2g\sin \theta }\\ H_A=\frac{u^2{\sin }^2\alpha }{2g\sin \theta }\\ H_C=\frac{u^2{\sin }^2\left(90-\alpha \right)}{2g\sin \theta }=\frac{u^2{\cos }^2\alpha }{2g\sin \theta }\\ \Rightarrow H_C+H_A=H_B\end{array}$

Hence,

option $\left(A\right)$ is correct answer.