Question

Two projectile thrown at $30^{\circ}$ and $45^{\circ}$ with the horizontal respectively, reach the maximum height in same time The ratio of their initial velocities is

• A. $1: \sqrt{2}$
• B. $2: 1$
• C. $\sqrt{2}: 1$
• D. $1: 2$

Answer 1

The Correct Option is C

To find the ratio of the initial velocities of two projectiles thrown at angles of 30^{\circ} and 45^{\circ} with the horizontal, we first need to understand the condition that both projectiles reach the maximum height in the same time.

The time to reach maximum height for any projectile is given by the formula: t = \frac{u \sin \theta}{g}, where u is the initial velocity, \theta is the angle with the horizontal, and g is the acceleration due to gravity.

For the first projectile thrown at 30^{\circ}:

t_1 = \frac{u_1 \sin 30^{\circ}}{g} = \frac{u_1 \cdot 0.5}{g} = \frac{u_1}{2g}

For the second projectile thrown at 45^{\circ}:

t_2 = \frac{u_2 \sin 45^{\circ}}{g} = \frac{u_2 \cdot \frac{\sqrt{2}}{2}}{g} = \frac{u_2 \sqrt{2}}{2g}

According to the problem, the time to reach maximum height for both projectiles is the same, hence:

\frac{u_1}{2g} = \frac{u_2 \sqrt{2}}{2g}

We can cancel out 2g from both sides:

u_1 = u_2 \sqrt{2}

This implies that the ratio of their initial velocities is:

\frac{u_1}{u_2} = \sqrt{2}:1

Thus, the ratio of their initial velocities is \sqrt{2}:1.