JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: A projectile is projected at angle $30^{\circ}$ and is at the same height at times $3 \, \text{s}$ and $5 \, \text{s}$.

Find: The speed of projection $u$.

For projectile motion, the two times corresponding to the same height add up to the total time of flight.

$$T = 3 + 5 = 8 \, \text{s}$$

Now use the time of flight formula:

$$T = \frac{2u \sin \theta}{g}$$

Substituting $T = 8 \, \text{s}$, $\theta = 30^{\circ}$, and $g = 10 \, \text{m/s}^2$,

$$8 = \frac{2u \sin 30^{\circ}}{10}$$$$8 = \frac{2u \times \frac{1}{2}}{10}$$$$8 = \frac{u}{10}$$$$u = 80 \, \text{m/s}$$

Therefore, the speed of projection is $80 \, \text{m/s}$. The correct option is C.

Common mistakes

• Adding or subtracting the two given times incorrectly. For the same height in projectile motion, the two times are symmetric about the highest point, so they must be added to get the total time of flight. Use $T = t_1 + t_2$ here.

• Using $\sin 30^{\circ} = 30/180$ or another incorrect value. The correct trigonometric value is $\sin 30^{\circ} = \frac{1}{2}$.

• Confusing speed of projection with horizontal velocity. The formula for time of flight uses the initial vertical component $u\sin\theta$, so substitute the full speed $u$ into $T = \frac{2u\sin\theta}{g}$.