A car starts from rest and accelerates at 5 , | vedclass

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(D) At $t=4 \, s$,the velocity of the car (and thus the ball) in the horizontal direction is $v_x = u + at = 0 + 5 	imes 4 = 20 \, m/s$. After the ba

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

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(D) At $t=4 \, s$,the velocity of the car (and thus the ball) in the horizontal direction is $v_x = u + at = 0 + 5 	imes 4 = 20 \, m/s$. After the ball is dropped,it is only under the influence of gravity. At $t=6 \, s$,the time elapsed since the ball was dropped is $\Delta t = 6 - 4 = 2 \, s$. The horizontal velocity remains constant: $v_x = 20 \, m/s$. The vertical velocity at $\Delta t = 2 \, s$ is $v_y = u_y + g \Delta t = 0 + 10 	imes 2 = 20 \, m/s$. The resultant velocity is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 20^2} = 20 \sqrt{2} \, m/s$. Since the ball is in free fall,its acceleration is equal to the acceleration due to gravity,$a = g = 10 \, m/s^{2}$ downwards.

$A$ car starts from rest and accelerates at $5 \, m/s^{2}$. At $t=4 \, s$,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 $t=6 \, s$? (Take $g = 10 \, m/s^{2}$)

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