JEE Physics 2023 Question with Solution

Given: Both projectiles are thrown with the same speed $v = 40 \, \text{m/s}$ and have the same range. One angle is $60^\circ$ and $g = 10 \, \text{m/s}^2$.

Find: The sum of the maximum heights of the two projectiles.

For the same speed, equal ranges occur for complementary projection angles. Therefore, if one angle is $60^\circ$, the other angle is $30^\circ$.

For projectile motion, maximum height is

$$H = \frac{v^2 \sin^2 \theta}{2g}$$

So for the first projectile,

$$H_1 = \frac{40^2 \sin^2 60^\circ}{2 \times 10}$$

Using $\sin 60^\circ = \frac{\sqrt{3}}{2}$,

$$H_1 = \frac{1600 \times \left(\frac{\sqrt{3}}{2}\right)^2}{20} = 40 \, \text{m}$$

For the second projectile,

$$H_2 = \frac{40^2 \sin^2 30^\circ}{2 \times 10}$$

Using $\sin 30^\circ = \frac{1}{2}$,

$$H_2 = \frac{1600 \times \left(\frac{1}{2}\right)^2}{20} = 20 \, \text{m}$$

Hence,

$$H_{\text{total}} = H_1 + H_2 = 40 + 20 = 80 \, \text{m}$$

Therefore, the required sum of maximum heights is $80$.

Given: Two projectiles are projected with the same speed and same range. One angle is $60^\circ$.

Find: Sum of their maximum heights.

The range of a projectile is

$$R = \frac{v^2 \sin 2\theta}{g}$$

If two projectiles with the same speed have equal ranges, then

$$\sin 2\theta_1 = \sin 2\theta_2$$

This gives complementary angles for distinct projections, so

$$\theta_1 + \theta_2 = 90^\circ$$

Since one angle is $60^\circ$, the other is

$$30^\circ$$

Now use the maximum height formula

$$H = \frac{v^2 \sin^2 \theta}{2g}$$

Then,

$$H_1 = \frac{40^2 \sin^2 60^\circ}{20}, \qquad H_2 = \frac{40^2 \sin^2 30^\circ}{20}$$

Substituting the trigonometric values,

$$H_1 = 40 \, \text{m}, \qquad H_2 = 20 \, \text{m}$$

Thus the sum is

$$80 \, \text{m}$$

Therefore, the answer is $80$.