Question
A car starts from rest and accelerates uniformly with 2 m/s2. At t = 5s a stone is dropped out of the window 15 m high of the car. What is velocity and acceleration of stone at t = 5·5 s? ( g = 10m/s2).
Solution
Given,
Car starts from rest. That is,
Initial velocity = 0 m/s
Uniform acceleration of the car = 2 m/s2
Velocity of car at t = 5s = 10 m/s
When stone is dropped, the velocity of car and hence velocity of stone is 10 m/s in horizontal direction.
Since at 5s the stone loses the contact from car, therefore, the stone will move uniformly in horizontal direction and accelerate in vertically downward direction due to gravity.
At t = 5·5s or 0·5 s after the stone is dropped, the X and Y component of velocity are 
So, magnitude of velocity is given by, 
Acceleration of the stone at t=5.5 sec is the same as acceleration due to gravity.
That is, acceleration = 10 m/s2
And,
Angle of inclination with the horizontal, 
= 
i.e., 
