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

Question

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

Professor PC Thomas Classes · Answer

(a) : Given : $u_{x}=0, a_{x}=5 \mathrm{~m} / \mathrm{s}^{2}$ At $t=4 \mathrm{~s}$ ball is dropped $a_{y}=g=10 \mathrm{~m} / \mathrm{s}^{2}$\ Let $v_{x}$ is the velocity of ball at the time $t=4 \mathrm{~s}$.\ $v_{x}=u_{x}+a_{x} t=0+5 	imes 4=20 \mathrm{~m} / \mathrm{s}$\ $v_{y}$ is the velocity of ball after leaving at $t=6 \mathrm{~s}$\ $v_{y}=u_{y}+a_{y} t^{\prime}=0+g 	imes 2=20 \mathrm{~m} / \mathrm{s}$\ Resultant velocity of ball at $t=6 \mathrm{~s}$ $$ v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{20^{2}+20^{2}} \Rightarrow v=20 \sqrt{2} \mathrm{~m} / \mathrm{s} $$ Acceleration of the ball will be due to gravity i.e., $a=10 \mathrm{~m} / \mathrm{s}^{2}$.\