Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A ball is projected from point $A$ with velocity $20 \text{ m s}^{-1}$ at an angle $60^{\circ}$ to the horizontal direction. At the highest point $B$ of the path (as shown in figure), the velocity $v$ (in $\text{m s}^{-1}$ ) of the ball will be:

$20$

$10\sqrt{3}$

Zero

$10$

Step-by-Step Solution

Analyze Velocity Components: In projectile motion, the velocity vector can be resolved into horizontal ( $u_x$ ) and vertical ( $u_y$ ) components. The horizontal acceleration is zero ( $a_x = 0$ ), so the horizontal component of velocity remains constant throughout the flight.
Velocity at Highest Point: At the highest point of the trajectory, the vertical component of the velocity becomes zero ( $v_y = 0$ ). Therefore, the total velocity at the highest point is entirely horizontal. $v_{\text{top}} = u_x = u \cos \theta$
Calculation: Given initial velocity $u = 20 \text{ m s}^{-1}$ and angle $\theta = 60^{\circ}$ . $v_{\text{top}} = 20 \cos(60^{\circ})$ $v_{\text{top}} = 20 \times 0.5 = 10 \text{ m s}^{-1}$

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