JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 17)

Two projectiles 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 :
$$1: \sqrt{2}$$
$$2: 1$$
$$\sqrt{2}: 1$$
$$1: 2$$

Explanation

To solve this problem, we need to understand the relationship between the angles of projection, the initial velocities, and the time taken to reach maximum height for each projectile.

The formula to calculate the time to reach the maximum height is given by:

$$ t = \frac{u \sin \theta}{g} $$

where:

  • $$t$$ is the time to reach maximum height,
  • $$u$$ is the initial velocity,
  • $$\theta$$ is the angle of projection, and
  • $$g$$ is the acceleration due to gravity.

Given that the projectiles reach the maximum height in the same time, we can set up the following equation:

$$ \frac{u_1 \sin 30^\circ}{g} = \frac{u_2 \sin 45^\circ}{g} $$

Since $$\sin 30^\circ = \frac{1}{2}$$ and $$\sin 45^\circ = \frac{\sqrt{2}}{2}$$, the equation simplifies to:

$$ \frac{u_1 \cdot \frac{1}{2}}{g} = \frac{u_2 \cdot \frac{\sqrt{2}}{2}}{g} $$

Canceling out the common terms (i.e., $$g$$ and $$\frac{1}{2}$$), we get:

$$ u_1 = u_2 \sqrt{2} $$

Hence, the ratio of their initial velocities is:

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

Therefore, the correct answer is Option C: $$\sqrt{2}:1$$.

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