# Bouncing ball and acceleration

1. Sep 30, 2004

### SahinTC

Hey there... I'm studying for a test tomorrow and there's a problem I can't seem to figure out... I hope you all don't mind me asking here (hey, it's what the forum's for ), and I hope you all can see that I really did attempt to work the problem myself.

Now, when the ball hits the ground, it's VInst is equal to 14m/s. Newton's 3rd law states that at that instant, the same force will be applied upwards towards the ball.

We know that the acceleration of a falling object is 10m/s^2. Yes, I am aware this is simplified, and would prefer to use 9.81 myself, but the department head for physics at my school rather seems to like rounding it off to a nice number.

If the velocity on impact was 14m/s, and the acceleration is 10m/s^2, the object took 1.4 seconds to hit the ground. We know the average velocity (free fall to 14m/s) is 7m/s.

X = X0 * Vavg * t...
Plugging everything in, the dX is equal to 7m/s*1.4s, leaving us with a distance of 9.8, which is h. I'm not quite sure of the relevence of this, but I figure it wouldn't hurt to find it. :P.

If the ball comes into contact with the ground for 1/27 of a second, then seconds 1.40 to 1.44 is the time in which the ball is on the ground. This is where I am severely confused. What happens here? The ground exerts an equal force of impact on the ball (17N), but doesn't it have to be greater than the force of gravity for it to bounce back up? The weight of the ball is also 16.7N (1.7kg * 9.8m/s), but these two forces can't be the same... acceleration would be zero and the ball would just hit the ground and stop.

So... what happens?

Also, the physics book is... well... for the most part, absolute garbage. I was wondering if any of you could recommend a good physics book or resource, ranging from the mechanics of physics to quantum physics, preferably.

2. Sep 30, 2004

### ShawnD

F = mass * acceleration

$$F = \frac{m \Delta v}{t}$$

$$F = \frac{(1.7)(14 - (-14))}{\frac{1}{27}}$$

1.7 is the mass, the time is 1/27, and the change in velocity is 28. The ball hit the ground going 14m/s downwards, so say that was -14m/s. For the ball to reach the same height, it would have to leave the ground at the same speed, but in the opposite direction, +14,/s. Total change of velocity was 28m/s.

F = 1285.2N

If somebody sees a problem with what I'm saying. I would love to hear it.

3. Sep 30, 2004

### Staff: Mentor

Looks good to me, ShawnD. The only nitpick I would make is that the F you found is the net force on the ball. The average force exerted by the ground would be F + mg. (Which just makes the original answer even more wrong.)

4. Sep 30, 2004

### Chi Meson

IF you take the same variables and look at it from the "impulse-momentum theorem" it takes you to the same place: Ft = m (delta)v Just like ShawnD's formula, but in this formula the force is specifically the average force. Again the answer is 1285 N, but sig figs makes the answer 1300 N. If the book actually says that this answer is 758 N, it really is garbage.

5. Sep 30, 2004

### SahinTC

You were indeed correct, ShawnD, the answer supplied is wrong...

Now someone point me to a good physics textbook before my brain explodes. ><

6. Sep 30, 2004

### Pyrrhus

Resnick and Halliday's Physics book, or try Serway and Beichnner Physics for scientists and engineers.

7. Oct 1, 2004

### Chi Meson

Cutnell and Johnson have a good algebra-based text, published by Wiley. For calculus, I like Sanny & Moebs from William C Brown.