Question: 7: A cricket ball is thrown by a player at a speed of 20 \mathrm{~m} / \mathrm{s} in a direction 30^{\circ} above the horizontal.
The maximum height attained by the ball during its motion is:
\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)(1) 5 m
(2) 10 m
(3) 20 m
(4) 25 m
Answer: Option (1)
Explanation:
The maximum height in projectile motion depends on the vertical component of velocity.
The vertical component of initial velocity is given by u_y = u \sin \theta.
Here, u = 20 \mathrm{~m/s} and \theta = 30^\circ
so u_y = 20 \times \sin 30^\circ = 20 \times \frac{1}{2} = 10 \mathrm{~m/s}.
The formula for maximum height is H = \frac{u_y^2}{2g}.
Substituting the values, we get H = \frac{(10)^2}{2 \times 10} = \frac{100}{20} = 5 \mathrm{~m}.
Therefore, the maximum height attained by the ball is 5 m.