Question

Four cannon balls, '1', '2', '3' and '4' are fired from level ground. Cannonball '1' is fired at an angle of ${60}^{\circ }$ above the horizontal and follows the path shown.

Cannonballs '2' and '3' are fired at an angle of ${45}^{\circ }$ and '4' is fired at an angle of ${30}^{\circ }$ above the horizontal. Which cannon ball has the largest initial speed?

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D

4

Expressions for range and maximum height are given by $R$ = $\frac{u^{2}sin2\theta }{g}$; $H=\frac{u^{2}si{n}^2\theta }{2g}$

From figure '2', '4' have same range $u_2^{2}sin(2\times {45}^{\circ }$) = $u_4^{2}sin(2\times {30}^{\circ }$); $u_2^{2}sin{90}^{\circ }$ = $u_4^{2}sin{60}^{\circ }$

Thus $u_2^2\left(1\right)=u_4^2\left(\frac{\sqrt{3}}{2}\right)$; $u_4>u_2$

Now we compare '2' and '3', $(Range{)}_3<(Range{)}_2$

As projection angle is same $u_3<u_2$

Now we compare '1' and '2', maximum height achieved by balls are same $H_1=H_2$

$\frac{u_1^{2}si{n}^2{60}^{\circ }}{2g}=\frac{u_2^{2}si{n}^2{45}^{\circ }}{2g}$

$u_1$ = $u_2\times \sqrt{\frac{2}{3}}$ or $u_2>u_1$

Thus, $u_4>u_2$ and $u_2>u_1$ and $u_2>u_3$

Therefore $u_4$ is maximum.

Hence initial velocity of the ball '4' is maximum.