Question

A body has maximum range

R_{1}

when projected up the plane. The same body when projected down the inclined plane, it has maximum range

R_{2}

. Find the maximum horizontal range. Assume that equal speed of projection in each case and the body is projected onto the inclined plane in the line of the greatest slope.

Accepted Answer

Let $θ_{0}$ be the angle of inclined plane with horizontal and $α$ be angle of projection from inclined plane.

Then for upward projection from inclined plane, maximum range is

$R_{\max} = \frac{u^{2}}{g (1 + \sin θ_{0})} = R_{1} \dots (i)$

For downward projection from inclined plane, maximum range is

$R_{\max} = \frac{u^{2}}{g (1 - \sin θ_{0})} = R_{2} \dots (ii)$

For a projection on horizontal surface $(θ_{0} = 0 °)$ and $(α = 45 °)$ ,

Maximum range is $R_{\max} = \frac{u^{2}}{g} = R$ (say)

To establish a relation between $R, R_{1}$ and $R_{2}$ , we need to eliminate $θ_{0}$ .

From equations $(i)$ and $(ii)$ , we get,

$R_{1} = \frac{R}{1 + \sin θ} \Rightarrow \sin θ = \frac{R - R_{1}}{R_{1}}$ and

$R_{2} = \frac{R}{1 + - \sin θ} \Rightarrow \sin θ = \frac{R_{2} - R}{R_{2}}$

Equating both equations, we get,

$\frac{R - R_{1}}{R_{1}} = \frac{R_{2} - R}{R_{2}}$

Simplifying the above relation,

$\frac{2}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

Hence, the maximum horizontal range is $R = \frac{2 R_{1} R_{2}}{R_{1} + R_{2}}$ .

B M Sharma Solutions for Chapter: Kinematics II, Exercise 43: ILLUSTRATION 5.40

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics II, Exercise 43: ILLUSTRATION 5.40

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