# Trajectories of a thrown object (with bumps considered)

Assume that I stood at the boundary line of a football / baseball / cricket ground(without any turf/pitch/grass in it) - just a plain ground - like play ground. I have a ball that can bounce when thrown on this ground.

My question is - When I throw the ball at an angle to ground/horizontal plane, the ball takes a trajectory- which is very clear. However, after hitting the ground (that is at first bounce), it takes another trajectory and so on and so forth. Are all these trajectories same in their shape (or the underlying equation, describing the trajectory), except for the amplitude/ maximum height it can reach?

Pictorially - you can imagine many half circles, with diminishing upper halves forming the loci (strictly speaking - not, though), whose centres are collinear.

The trajectories of the thrown object and its consecutive bounces will be parabolic in their shape. In this case the ball will lose some of its kinetic energy each time it make contact with the Ground. However, the angle it makes with the ground each time will be equal to the angle it was original throw at. This means that all of the trajectories will be scaled down versions of one another, with the scale factor being the fraction of energy is keeps each time it bounces.

the position of the particle as a function of time is

x=Vcos(theta)*t

y= Vsin(theta)T-0.5gt^2

where v is the initial velocity and theta is the initial launch angle (g is 9.8 m/s^2). You can solve for the trajectory it terms of x y by solving for t in the first equation and substituting into the second equation.

Then all of the other parabolic trajectories will be the same parabola of form y=ax^2+bx+c, however, the a value will be multiplied by E2/E1, where e2 is the kinetic energy of the ball after a given collision and E1 is the kinetic energy of the ball just before that same collision. This numbers cant be determined without knowing how inelastic or elastic the collision is. However, the trajectories will be the same parabolic trajectory, just scaled down by the factor E2/E1 (of course it will also be moved to the left or right that the x intercept of one trajectory, I.E, where it leaves the ground, lines up with where it hit the ground, I.E the x intercept of on of the other trajectories.)

The trajectories will also be restricted to have only a positive range because the ball will never go below the ground in a normal situation like you describe.

If you sole the above equations the initial trajectory would be:

y=tan(theta)x-(gx^2)/(2v^2*cos^2(theta))

• Well, it answers my question. If possible, please provide equations which express these trajectories, with diminishing amplitudes, such as - y1 = A1 Sin x t ( for the initial one), y2 = A2 Sin xt.,.... . Thanks. – Mea Culpa Nay Sep 8 '19 at 15:27