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# It is possible to project a particle with a given velocity in two possible ways so as to make it pass through a point at a horizontal distance R from the point of projection .If $H_1$ and $H_2$ are the corresponding maximum height and $H$ is maximum height when the particle is projected at $45^{\circ}$ with the same speed, then what is $H_1+H_2$

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When a particle is projected at an angle $\theta_0$ and again at an angle $(90^{\circ})$ with the same velocity , then although the range R is same in both cases yet the maximum heights are different . If these are $H_1$ and $H_2$ respectively then,
$H_1 =\large\frac{v_0^2 \sin ^2 \theta_0}{2g}$-----(i)
$H_2 = \large\frac{v_0^2 \sin ^2 (90^{\circ}-\theta_0)}{2g}$-----(ii)
Adding (i) and (ii)
$H_1 +H_2 = \large\frac{v_0^2}{2g}$-----(iii)
The maximum height to which a projectile can be projected with velocity $v_0$ when $\theta=45^{\circ}$ is given by,
$H=\large\frac{v_0^2 \sin ^2 45^{\circ}}{2g} = \large\frac{v_0^2}{2(2g)}$------(iv)
From (iii) and (iv)
$H_1+H_2= 2H$
answered Jul 15, 2014 by
edited Aug 13, 2014

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