Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A car starts from rest and accelerates at $5 \text{ m/s}^2$ . At $t=4 \text{ s}$ , a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t=6 \text{ s}$ ? (Take $g=10 \text{ m/s}^2$ )

$20\sqrt{2} \text{ m/s}, 0 \text{ m/s}^2$

$20\sqrt{2} \text{ m/s}, 10 \text{ m/s}^2$

$20 \text{ m/s}, 5 \text{ m/s}^2$

$20 \text{ m/s}, 0 \text{ m/s}^2$

Step-by-Step Solution

Analyze Motion of Car (t = 0 to 4 s): The car accelerates from rest ( $u=0$ ) with $a = 5 \text{ m/s}^2$ . Velocity of the car at $t = 4 \text{ s}$ is: $v_{car} = u + at = 0 + 5(4) = 20 \text{ m/s}$
Analyze Motion of Ball (t > 4 s): At $t=4 \text{ s}$ , the ball is dropped. Due to inertia, it acquires the horizontal velocity of the car at that instant. The ball becomes a projectile.

Horizontal component ( $v_x$ ): Remains constant (neglecting air resistance). $v_x = 20 \text{ m/s}$
Vertical component ( $v_y$ ): Initial vertical velocity $u_y = 0$ (dropped). The ball falls under gravity ( $g = 10 \text{ m/s}^2$ ). Time elapsed since dropping $\Delta t = 6 \text{ s} - 4 \text{ s} = 2 \text{ s}$ . $v_y = u_y + g(\Delta t) = 0 + 10(2) = 20 \text{ m/s}$

Calculate Resultant Velocity: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = 20\sqrt{2} \text{ m/s}$
Calculate Acceleration: Once the ball is released, the only force acting on it is gravity. Therefore, the acceleration is constant and equal to $g$ acting vertically downwards. $a = 10 \text{ m/s}^2$

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