Question

A ball is thrown at different angles with the same speed $u$ and from the same point and it has the same range in both the cases. If $y_1$ and $y_2$ are the heights attained in the two cases, then $y_1+y_2$ is equal to

• A. $\frac{u^2}{g}$
• B. $\frac{2u^2}{g}$
• C. $\frac{u^2}{2g}$
• D. $\frac{u^2}{4g}$

Answer 1

The correct option is C $\frac{u^2}{2g}$

ranges of both projectile motion are equal therefore

$R_1=R_2$

$\frac{v_o^{2}sin2{\theta }_1}{g}=\frac{v_o^{2}sin2{\theta }_2}{g}$

$sin2{\theta }_1=sin2{\theta }_2$ --------------$\left(i\right)$

$2{\theta }_1=\pi -2{\theta }_2$

${\theta }_1=\frac{\pi }{2}-{\theta }_2$

maximum height of first projectile at angle $\frac{\pi }{3}$ is $y_1$

$y_1=\frac{v_o^{2}si{n}^2{\theta }_1}{2g}$

$y_2=\frac{v_o^{2}si{n}^2{\theta }_2}{2g}$

therefore $y_1+y_2=\frac{v_o^2}{2g}(si{n}^2{\theta }_1+si{n}^2{\theta }_2)=\frac{v_o^2}{2g}(si{n}^2{\theta }_1+si{n}^2(\frac{\pi }{2}-{\theta }_1\left)\right)=\frac{v_o^2}{2g}(si{n}^2{\theta }_1+co{s}^2{\theta }_1)=\frac{v_o^2}{2g}$