A projectile is fired from the ground with an initial speed of 50 m/s at an angle of 30^ with the horizontal. The maximum height attained by the projectile is:

A projectile is fired from the ground with an initial speed of 50… — Solution | JEE Physics

MCQEasyJEE 2024Projectile Motion

A projectile is fired from the ground with an initial speed of $50 \, \text{m/s}$ at an angle of $30^\circ$ with the horizontal. The maximum height attained by the projectile is:

A
$31.25 \, \text{m}$
B
$25 \, \text{m}$
C
$50 \, \text{m}$
D
$37.5 \, \text{m}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Initial speed is $u = 50 \, \text{m/s}$ and angle of projection is $\theta = 30^\circ$ .

Find: The maximum height attained by the projectile.

The solution states that the correct option is A, but the displayed working is for a different projectile question involving circular motion. For this question, the relevant projectile formula is:

H = \frac{u^2 \sin^2\theta}{2g}

Using $g = 10 \, \text{m/s}^2$ and $\sin 30^\circ = \frac{1}{2}$ ,

H = \frac{(50)^2 \left(\frac{1}{2}\right)^2}{2 \times 10}

H = \frac{2500 \times \frac{1}{4}}{20}

H = \frac{625}{20} = 31.25 \, \text{m}

Therefore, the maximum height attained by the projectile is $31.25 \, \text{m}$ . The correct option is A.

Common mistakes

Using the formula for range instead of maximum height is incorrect because range depends on $\sin 2\theta$ , while height depends on the vertical component only. Use $H = \frac{u^2 \sin^2\theta}{2g}$ .
Taking $\sin 30^\circ = 1$ or forgetting to square it gives a wrong value. Here $\sin 30^\circ = \frac{1}{2}$ , so $\sin^2 30^\circ = \frac{1}{4}$ .
Substituting $g = 9.8 \, \text{m/s}^2$ without checking the exam convention can lead to an option mismatch. In standard JEE-style options like these, use $g = 10 \, \text{m/s}^2$ unless specified otherwise.

Practice more Projectile Motion questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Projectile Motion

Answer

Step-by-step solution

Common mistakes

Practice more Projectile Motion questions

Related questions