Question

A ball is thrown at a speed of 20 m $s^{-1}$ at an angle of $30^{0}$ with the horizontal. The maximum height reached by the ball is ($g=10~ms^{-2}$)}

• A. 2 m
• B. 3 m
• C. 4 m
• D. 5 m

Hint:

Vertical component of velocity is $u \sin \theta$. Height depends only on this vertical component ($v_{y}^{2}/2g$).

Answer 1

The Correct Option is D

Step 1: Concept
For a projectile, the maximum height $H$ is given by $H = \frac{u^{2} \sin^{2} \theta}{2g}$.

Step 2: Meaning
Given $u = 20$ m/s, $\theta = 30^{\circ}$, and $g = 10$ m/$s^{2}$.

Step 3: Analysis
$\sin 30^{\circ} = 1/2$. Therefore, $\sin^{2} 30^{\circ} = 1/4$.

Step 4: Conclusion
$H = \frac{20^{2} \times (1/4)}{2 \times 10} = \frac{400 \times 0.25}{20} = \frac{100}{20} = 5$ m. Final Answer: (D)