# Is the final velocity of a Tossed tomato the same as its initial velocity

• B
• rudransh verma
Rudransh Verma.In summary, the energy-conservation law states that the total energy of a particle in the gravitational field of Earth is constant.f

#### rudransh verma

Gold Member
When a tomato is thrown up with a velocity ##v_0## it’s kinetic energy is 1/2mv_0^2. It will stop at the top and then again comes back to the launch point where it’s kinetic energy will be same as before, 1/2mv_0^2. How is this possible?

Also when we throw the tomato up how can you be so sure that it’s the tomato that is rising and not that the Earth is going down away from the tomato ?

That's the energy-conservation law (see the other thread on the work-energy theorem). It says that for the motion of a particle (take the tomato as a "particle" here) in the gravitational field of the Earth (approximated as constant) the total energy
$$E=E_{\text{kin}}+V=\frac{m}{2} \vec{v}^2 + m g z=\text{const}.$$
First of all if the tomato starts at ##(x,y,z)=(0,0,0)## with velocity ##\vec{v}_0=(0,0,v_0)## these initial conditions tell you that
$$E=\frac{m}{2} v_0^2 = \frac{m}{2} v^2 + m g z=\text{const}.$$
This tells you that the maximum value ##z_{\text{max}}=h## is reached when ##v=0##, i.e., for
$$\frac{m}{2} v_0^2 = m g h \; \Rightarrow \; h=\frac{v_0^2}{2 g}.$$
At this point the particle must fall down again. If it again reaches ##z=0##, it's velocity must again be ##v_0##.

Note that all this you can read off the energy-conservation law without solving the equations of motion.

If we ignore wind resistance, yes..
When a tomato is thrown up with a velocity ##v_0## it’s kinetic energy is 1/2mv_0^2. It will stop at the top and then again comes back to the launch point where it’s kinetic energy will be same as before, 1/2mv_0^2. How is this possible?
Because the math says so? Because parabolas are clearly symmetrical? Because God was too lazy to make it more complicated?

I don't know what sort of answer will satisfy a question like "how is this possible?" when you have the math right in front of you.
Also when we throw the tomato up how can you be so sure that it’s the tomato that is rising and not that the Earth is going down away from the tomato ?
That can get complicated but the easy answer is that the tomato had a force applied so by f=ma we know it accelerated. If you want you can apply that to Earth and see if it makes a difference to ignore it.

• nasu and vanhees71
That's the energy-conservation law (see the other thread on the work-energy theorem). It says that for the motion of a particle (take the tomato as a "particle" here) in the gravitational field of the Earth (approximated as constant) the total energy
$$E=E_{\text{kin}}+V=\frac{m}{2} \vec{v}^2 + m g z=\text{const}.$$
First of all if the tomato starts at ##(x,y,z)=(0,0,0)## with velocity ##\vec{v}_0=(0,0,v_0)## these initial conditions tell you that
$$E=\frac{m}{2} v_0^2 = \frac{m}{2} v^2 + m g z=\text{const}.$$
This tells you that the maximum value ##z_{\text{max}}=h## is reached when ##v=0##, i.e., for
$$\frac{m}{2} v_0^2 = m g h \; \Rightarrow \; h=\frac{v_0^2}{2 g}.$$
At this point the particle must fall down again. If it again reaches ##z=0##, it's velocity must again be ##v_0##.

Note that all this you can read off the energy-conservation law without solving the equations of motion.
So your second eqn says a lot of things like the initial kinetic energy changes to sum of potential and kinetic energies at some other instant and this sum is always constant. For max h we need v=0. At this point all of the initial kinetic energy changes to potential energy. With this energy it comes down and at z=0 v=v_0.

• vanhees71
If you want you can apply that to Earth and see if it makes a difference to ignore it.
@rudransh verma : This is worth doing at least one or two times. Especially for a student who doesn't like to take things for granted just because someone says so. You will need to know, the Earth's mass is 6*10^24 kg, and you can pick a mass for the tomato (0.125 kg maybe?).

• rudransh verma, vanhees71 and russ_watters
@rudransh verma : This is worth doing at least one or two times. Especially for a student who doesn't like to take things for granted just because someone says so.
For the second trial I suggest the Moon instead of a tomato.

I have a better idea. See my post
Thread 'To displace Earth from its orbit'
OK, but I don't see where you have worked out the answer to your question there. The point of my suggestion is that YOU work it out, that's how each of us has learned physics.

• vanhees71 and russ_watters
How is this possible?
Conservation of energy as explained by @vanhees71 . Don't forget that you have gravity (a force) that acts constantly on the tomato, hence the deceleration on the way up followed by an acceleration in the opposite direction.
Also when we throw the tomato up how can you be so sure that it’s the tomato that is rising and not that the Earth is going down away from the tomato ?
Conservation of momentum together with conservation of energy. In the following problem, replace one cart with your tomato and the other one with the Earth and find out the velocity of each one:

### Is the final velocity of a Tossed tomato the same as its initial velocity​

Commenting on the title instead of the posts:

The final velocity would be the same as the initial velocity only if the net momentum transferred to the tomato over the duration of the flight was zero. This is usually not the case with a tossed tomato, so while the speed might be similar if there had been no wind resistance (conservation of energy and all) and net altitude change, the velocity would tend more towards downward instead of its initial upward trajectory.

Also when we throw the tomato up how can you be so sure that it’s the tomato that is rising and not that the Earth is going down away from the tomato ?
Actually, the Earth does go down and back up a bit. Conservation of momentum demands it.

Last edited:
Actually the speeds would be identical, with all the caveats from the above posts. Velocity is a vector quantity.