Check this question first please


My question is: Why can't I use equations of motion to get the final speed after rebounding?Acceleration not equal to $9.8 \text{m/s}^2$ or the intial after rebounding doesnt equal the intial before? Why do I have to use $$mgh=\frac{1}{2} mv^2$$ $$\text{Total mechanical before}=\text{after}$$

  • $\begingroup$ You can. You could use either conservation of energy or kinematics to find the speed. Does that answer your question? $\endgroup$ – Brionius Jan 5 '15 at 3:02
  • $\begingroup$ Some of the energy was absorb by the floor, otherwise it would bounce up to 1.25m. $\endgroup$ – LDC3 Jan 5 '15 at 3:19
  • $\begingroup$ Don't paste photos of a problem. Take the time to write it yourself. By asking us to read that photo you are not respecting the other users. Your title also could use work. Please see this post. $\endgroup$ – DanielSank Jan 5 '15 at 8:59

You can use either method, conservation of energy or kinematics, and both will give the same answer:


Let's arbitrarily define downwards displacements, velocities, accelerations and forces as negative.

Falling motion:

$$mgh_1 = \frac{1}{2}mv_1^2$$

$$\therefore v_1 = -\sqrt{2gh_1}$$

Rising motion:

$$\frac{1}{2}mv_2^2 = mgh_2$$

$$\therefore v_2 = \sqrt{2gh_2}$$

$$I = m(v_2 - v_1) = m(\sqrt{2gh_2}+\sqrt{2gh_1})$$


Constant acceleration, therefore we can use suvat equations. $$v^2 = u^2 +2as$$


$$v_1^2 = 0 + 2(-g)(-h_1)$$

$$\therefore v_1 = -\sqrt{2gh_1}$$


$$0 = v_2^2 + 2(-g)(h_2)$$

$$\therefore v_2 = \sqrt{2gh_2}$$

$$I = m(v_2 - v_1) = m(\sqrt{2gh_2}+\sqrt{2gh_1})$$

The reason why you get the same answers either way is because the definition of KE derived from kinematics.

The kinetic energy is the amount of work require to bring a mass, $m$ to a velocity $v$ from rest.

$$KE = work = Fs = mas$$

suvat equation:

$$v^2 = u^2 +2as$$

$$\therefore v^2 = 0 + 2as$$

Substitute: $$v^2 = \frac{2KE}{m}$$

$$\therefore KE = \frac{1}{2}mv^2$$

  • $\begingroup$ In the conservation of energy part why the falling motion is in negative and the rising in postive?? $\endgroup$ – Ahmed Alkhatib Jan 5 '15 at 12:20
  • $\begingroup$ That is because, whenever you take the square root of both sides, there is an ambiguity to whether one of the sides are positive or negative. So there was a choice to make the velocity positive or negative. However, I defined a convention in the question that states that downward velocities are negative. Therefore, in the case of the falling motion, the velocity was downwards, so, intuitively, it had to be negative, hence the minus sign. This applies similarly to the rising motion case: velocity is upwards, therefore I chose it to be positive. $\endgroup$ – Involutius Jan 5 '15 at 12:27

Simple answer: you have to use the information you are given. I am not quite sure what equation of motion you intend to use - you have very little information about the interaction at the floor, only that the object apparently lost some energy (which is why it ends up reaching a lower height after the bounce).

The simplest way to do this is with conservation of energy - but equally you could "play time in reverse" and start with the ball at 0.96 m, drop it, and see what its speed was just as it hit the floor. Compare this with the speed it had when it reached the floor in "normal" time - after dropping 1.25 m. The total impulse given is the impulse needed to effect that change in direction.

That's "almost" using equations of motion. The mechanics is the same - how you choose to solve it is a choice you get to make.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.