Question

A particle is projected in air at an angle $\beta$to a surface which itself is inclined at an angle $\alpha$to the horizontal. Then distance L is equal to

• A. $\frac{2u^{2}\sin\alpha\cos(\alpha + \beta)}{g\cos^{2}\alpha}$
• B. $\frac{2u^{2}\sin\beta\cos(\alpha + \beta)}{g\cos^{2}\beta}$
• C. $\frac{2u^{2}\sin\beta\cos(\alpha + \beta)}{g\cos^{2}\alpha}$
• D. $\frac{2u^{2}\sin\alpha\cos(\alpha + \beta)}{g\cos^{2}\beta}$

Answer 1

Answer: (C)

$\frac{2u^{2}\sin\beta\cos(\alpha + \beta)}{g\cos^{2}\alpha}$

Answer 2

Take the x – axis along the incline and y – axis perpendicular to the plane.

$\therefore u_{s} = u\cos\beta$

$u_{y} = u\sin\beta$

$a_{x} = - g\sin\alpha$

$a_{y} = - g\cos\alpha$

When the particle lands at P its y coordinate becomes zero.

$\therefore 0 = u_{y}t + \frac{1}{2}a_{y}t^{2}$

$0 = u\sin\beta t - \frac{1}{2}g\cos\alpha t^{2}$

Or $t = \frac{2u\sin\beta}{g\cos\alpha}$ ….. (i)

For motions along inclined plane

$x = u\sin\beta t - \frac{1}{2}g\cos\alpha t^{2}$

$\therefore L = u\cos\beta t = \frac{1}{2}g\sin\alpha t_{2}$

Substituting the value of t form Eq. (i) we get

$L = u\cos\beta\left( \frac{2u\sin\beta}{g\cos\alpha} \right) - \frac{1}{2}g\sin\alpha\left( \frac{2u\sin\beta}{g\cos\alpha} \right)$

$= \frac{2u^{2}\sin\beta\cos\beta}{g\cos\alpha} - \frac{2u^{2}\sin\alpha\sin^{2}\beta}{g\cos^{2}\alpha}$

$= \frac{2u^{2}}{g\cos^{2}\alpha}\lbrack\sin\beta\cos\beta\cos\alpha - \sin\alpha\sin^{2}\beta\rbrack = \frac{2u^{2}\sin\beta}{g\cos^{2}\alpha}\lbrack\cos\beta\cos\alpha - \sin\alpha\sin\beta\rbrack = \frac{2u^{2}\sin\beta\cos(\alpha + \beta)}{g\cos^{2}\alpha}$