This seems easy ..... The ball is thrown outside the window ......... this means it will have the horizontal velocity that of the car at that instant ....... and no vertical velocity initially ..... Then, there even collisions can occur as the time is too large ......
1. First find out the velocity of the car at t=10 sec using the formula
v = u + at [Here u = 0]
Due to inertia this velocity will be the velocity of the stone.
2. When the stone is projected from the car then it only experiences an acc. due to gravity ie g. So no matter what be the time the acc. of the stone will be g. If after 10.1 sec from projection the stone reaches the ground then the wt of the stone will be balannced by the normal rxn provided by the ground on the stone and net acc will be 0, otherwise it will fall freely with an acc of g. I think the question is on finding the velocity of the stone after 10.1 sec of projection. If so then find out the vertical velocity of the stone after 10.1 sec(say z), and also check whether it will reach the ground or not. If it reaches the ground then the velocity will be 0. If not then the resultant velocity will be the resulant of the vectors v and z. Also find out the angle of the resultant velocity with using the formula of the vector...
These are the concepts. If u want the soln. then nudge me!!!!!!!!!!!!!
Whenever u feel bad go for math
if u feel too bad
imagine your rival competeing u
U will be energetic like never before
Ya, the magnitude of accleration will be always be g .......
But, as the time period is too long ........ i think the ball will keep bouncing ...... not to the same height ( as it is not perfectly elastic collision) ...... but surely it is not perfectly inelastic collision ......
Its so simple yaaar....... At 10.1 s the ball wiil have horizontal velocity of car at t=10s and a vertical velocity downward g/10 which it has get in 0.1 s and off course its acceleration will be g with respect to earth and square root of (g sq. + a sq. ) with respect to car