Question
A LIFT IS MOVING UPWARD WITH ACC 2m/s2 .WHEN IT GAIN THE VELOCITY 4m/s THEN AT THAT MOMENT A BALL IS PROJECTED BY 300 (wrt floor to lift) WITH VEL. 4m/s RELATIVE TO LIFT . FIND HOW MUCH TIME WILL IT TAKE TO 1) TO REACH GROUND & 2) TO THE FLOOR OF LIFT AGAIN
Joshi sir comment
since velocity of the ball is 4 m/s relative to lift so
time taken by the ball to reach the floor of lift T = 2usinθ/g = (2*4*1/2)/10 = 0.4 sec
height of the lift at the time of dropping the ball = (v2-u2)/2g = 16/20 = 8/10 = 0.8 m = h
and vertical component of the velocity of ball with respect to the ground = 6 m/s = u
so use -h = ut - gt2/2