Question

a projectile can have same range from two angles of projection with same initial speed .If H1 and H2 be the maximum height then​

Answer 1

Let the angle of projection for the projectile in the two projections be $\sf{\theta_1}$ and $\sf{\theta_2}$ respectively, where $\sf{\theta_1\neq\theta_2.}$ Let the initial speed be u which is same in both cases.

The horizontal range of projectile is,

$\displaystyle\longrightarrow\sf{R=\dfrac{u^2\sin(2\theta)}{g}}$

Ranges in both cases are same.

$\displaystyle\longrightarrow\sf{R_1=R_2}$

$\displaystyle\longrightarrow\sf{\dfrac{u^2\sin(2\theta_1)}{g}=\dfrac{u^2\sin(2\theta_2)}{g}}$

$\displaystyle\longrightarrow\sf{\sin(2\theta_1)=\sin(2\theta_2)}$

Since $\sf{\theta_1\neq\theta_2,}$ possibly,

$\displaystyle\longrightarrow\sf{2\theta_1+2\theta_2=180^o}$

Because we know that $\sf{\sin(180^o-\theta)=\sin\theta.}$

$\displaystyle\longrightarrow\sf{\theta_1+\theta_2=90^o}$

$\displaystyle\longrightarrow\sf{\theta_2=90^o-\theta_1\quad\quad\dots(1)}$

The maximum height of the projectile in the first case is,

$\displaystyle\longrightarrow\sf{H_1=\dfrac{u^2\sin^2\theta_1}{2g}\quad\quad\dots(2)}$

And that in the second case is,

$\displaystyle\longrightarrow\sf{H_2=\dfrac{u^2\sin^2\theta_2}{2g}\quad\quad\dots(3)}$

Dividing (2) by (3),

$\displaystyle\longrightarrow\sf{\dfrac{H_1}{H_2}=\dfrac{\dfrac{u^2\sin^2\theta_1}{2g}}{\dfrac{u^2\sin^2\theta_2}{2g}}}$

$\displaystyle\longrightarrow\sf{\dfrac{H_1}{H_2}=\dfrac{\sin^2\theta_1}{\sin^2\theta_2}}$

From (1),

$\displaystyle\longrightarrow\sf{\dfrac{H_1}{H_2}=\dfrac{\sin^2\theta_1}{\sin^2(90^o-\theta_1)}}$

Since $\sf{\sin(90^o-\theta)=\cos\theta,}$

$\displaystyle\longrightarrow\sf{\dfrac{H_1}{H_2}=\dfrac{\sin^2\theta_1}{\cos^2\theta_1}}$

$\displaystyle\longrightarrow\sf{\dfrac{H_1}{H_2}=\tan^2\theta_1}$

$\displaystyle\longrightarrow\sf{\underline{\underline{H_1=H_2\tan^2\theta_1}}}$

In terms of $\sf{\theta_2,}$

$\displaystyle\longrightarrow\sf{H_1=H_2\tan^2(90^o-\theta_2)}$

Since $\sf{\tan(90^o-\theta)=\cot\theta,}$

$\displaystyle\longrightarrow\sf{\underline{\underline{H_2=H_1\tan^2\theta_2}}}$