Question

Two stone are projected with the same speed but making diffrent angles with the horizontal. Their ranges are equal. If the angle of projection of one is $\frac{\pi }{3}$ and its maximum height is $y_1$ then the maximum height of the other will be:

• A. $3y_1$
• B. $2y_1$
• C. $\frac{y_1}{2}$
• D. $\frac{y_1}{3}$

Answer 1

Answer: (D)

$\frac{y_1}{3}$

ranges of both projectile motion are equal therefore

$R_1=R_2$

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

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

$\Rightarrow {\theta }_1+{\theta }_2=\frac{\pi }{2}$

$\Rightarrow {\theta }_2=\frac{\pi }{6}$

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

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

$y_1=\frac{v_o^{2}si{n}^2\frac{\pi }{3}}{2g}=\frac{3v_o^2}{8g}$

$y_2=\frac{v_o^{2}si{n}_2^{\theta }}{2g}=\frac{v_0^2\sin {30}^0}{2g}=\frac{v_0^2}{8g}$

$\therefore y_2=\frac{y_1}{3}$