A cricket ball is thrown by a player at a speed of $20 ext{ m/s}$ in a direction $30^{\circ}$ above the hori

Question

A cricket ball is thrown by a player at a speed of $20 	ext{ m/s}$ in a direction $30^{\circ}$ above the horizontal. The maximum height attained by the ball during its motion is: (Take $g=10 	ext{ m/s}^2$)

Accepted Answer

The maximum height $h_m$ reached by a projectile is given by the formula:
$$ h_m = \frac{(v_0 \sin 	heta_0)^2}{2g} $$
where $v_0$ is the initial speed, $	heta_0$ is the angle of projection, and $g$ is the acceleration due to gravity .

Given:
*   Initial speed, $v_0 = 20 	ext{ m/s}$
*   Angle of projection, $	heta_0 = 30^{\circ}$
*   Acceleration due to gravity, $g = 10 	ext{ m/s}^2$

Substituting these values into the equation:
$$ h_m = \frac{(20 \sin 30^{\circ})^2}{2 	imes 10} $$
$$ h_m = \frac{(20 	imes 0.5)^2}{20} $$
$$ h_m = \frac{(10)^2}{20} $$
$$ h_m = \frac{100}{20} = 5 	ext{ m} $$

Thus, the maximum height attained by the ball is $5 	ext{ m}$.

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