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I have a ball, and I throw it at a wall as a projectile (assume that the ball's position the instant the journey begins is on level-ground). The ball hits the wall with a velocity that is at right angles to the wall, and bounces off, landing a distance closer to the wall relative to the launch position.

I know the angle of launch, as well as the initial velocity (thus both horizontal and vertical components), and the distance of the initial position as well as the final position from the wall.

Using this, I can calculate the height of impact (by using the SUVAT equations and letting $v_{final} = 0$) as "The ball hits the wall with a velocity that is at right angles to the wall", as well as the time taken to get to the wall.

I want to know the instantaneous velocity immediately after the ball rebounds off the wall - my friend believes it is the equal to the distance between the landing position and the wall divided by the time taken to get to the wall (as calculated before, the time from launch to impact), but I am not convinced.

Is my friend right? If so, why? Is this even enough information to determine the instantaneous velocity after impact?

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  • $\begingroup$ i think one can consider the return trajectory and can get the velocity using the distance from the wall of landing point and time taken. $\endgroup$ – drvrm Mar 28 '16 at 10:30
  • $\begingroup$ What is SUVAT? Please define acronyms unless you are sure that they will be understood. (I can't think of too many of those.) $\endgroup$ – garyp Mar 28 '16 at 15:50
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Your friend is right. The time of flight to the wall and back will be the same - you should be able to show this by analyzing the vertical motion for each. Assuming air resistance is not a factor the horizontal component of the ball's velocity will remain constant during its return flight so dividing the horizontal distance back by the time gives the average horizontal velocity. Since the horizontal velocity is constant this average velocity is the instantaneous velocity right after impact.

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  • $\begingroup$ The time of flight to the wall and back will be the same. According to the conditions stated in the question (the ball lands closer to the wall after bouncing off it) this is not true. It appears that the collision is not totally elastic. $\endgroup$ – sammy gerbil Jul 31 '18 at 11:06
  • $\begingroup$ True, the collision is not totally elastic or the motion would be symmetrical with the ball landing back where it was launched. The ball's horizontal speed on the return flight is smaller than it horizontal speed on the initial flight and accordingly the horizontal distance it travels or the return flight is less. We know the time of flight for each flight is the same from the vertical motion which is symmetrical. $\endgroup$ – M. Enns Jul 31 '18 at 14:33
  • $\begingroup$ My apologies, you are correct. $\endgroup$ – sammy gerbil Jul 31 '18 at 14:36

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