A car starts from rest and accelerates at $4 \mathrm{~m} / \mathrm{s}^{2}$. At $t=5 \mathrm{~s}$, a

Question

A car starts from rest and accelerates at $4 \mathrm{~m} / \mathrm{s}^{2}$. At $t=5 \mathrm{~s}$, a ball is dropped out of a window by a person sitting in the car. What is the velocity of the ball at $t=8 \mathrm{~s}$? (Take $g=10 \mathrm{~m} / \mathrm{s}^{2}$)

Professor PC Thomas Classes · Answer

(a) : At $t=5 \mathrm{~s}$, the car's horizontal velocity is $v_x = u + at = 0 + 4 	imes 5 = 20 \mathrm{~m} / \mathrm{s}$. \ When dropped, the ball retains this horizontal velocity: $v_x = 20 \mathrm{~m} / \mathrm{s}$. \ The vertical velocity after $8-5 = 3 \mathrm{~s}$ of free fall is $v_y = gt = 10 	imes 3 = 30 \mathrm{~m} / \mathrm{s}$. \ Resultant velocity $v = \sqrt{20^2 + 30^2} = \sqrt{400+900} = \sqrt{1300} = 10\sqrt{13} \mathrm{~m} / \mathrm{s}$.