# Initial Velocity Question

What is the initial velocity vertically when a person jumps 0.5 metres into the air, neglecting horizontal movement. Is it just 0 m/s? And the final velocity would be -9.8m/s^2 * the time it took took to land after reaching the max height?

Edit- Isn't the final vertical velocity 0? Because d2-d1/t2-t1 = v, thus there is no change in displacement because you are back where you vertically started.

## 2 Answers

Knowing the final velocity $v_f = 0$ $m/s$, the acceleration is $a=-9.8$ $m/s^2$, and that the displacement is $\Delta x = 0.5$ $m$, you can use the kinematic equation $v_f^2 = v_i^2 + 2a\Delta x$ to find the initial velocity to be approximately $3.13$ $m/s$. Although the person starts at rest, the initial velocity is not zero, since they have to jump - and therefore accelerate against Earth's gravitational field. The kinematic equations assume constant acceleration, so the "initial velocity" is actually referring to the speed just after their feet leave the ground.

The final velocity is also not zero, for the same reason. Final velocity in this context will mean the speed just before they land. It's all a consequence of having to assume constant acceleration.

Here I'll neglect air resistance (as well as assume there is no horizontal movement, as you say). Additionally, I'll say 'velocity' rather than 'speed', but since we're only moving in one dimension the velocity differs from the speed only by a sign.

The initial velocity is provided by the act of jumping, or more precisely of pushing off the ground beneath, which we will assume to be an essentially instantaneous process (for simplicity). This gives the person some initial velocity $u$, which points directly upwards (away from the ground). Until the person again strikes the ground (upon landing), the acceleration is constant and provided entirely by a gravitational force, directed exactly downwards.

You seem to be using the 'SUVAT' equations, specifically $v=u+at$ (where $u$ is the initial velocity, $v$ the final velocity, $a$ the acceleration, and $t$ the time) to calculate the final velocity $v$. This is correct, and if you use the correct initial velocity $u$ you will find that just as the person reaches the ground (but before they actually strike the ground) their velocity is exactly minus the initial velocity (i.e. the magnitude of the velocity is as it was initially, but the direction is reversed). Note that to use the SUVAT equations there must be a constant acceleration, which is the case here (after leaving the ground and before striking it again).

In order to calculate the required initial velocity to reach a certain height $h$, you can use one of the SUVAT equations, specifically for the period between jumping and reaching the maximum height. This is the kinematic equation that zhutchens1 uses in their answer. This is equivalent to a use of conservation of energy, to say that $\tfrac{1}{2}mu^2=mah$ where $h$ is the height reached, $m$ is the mass of the person (which cancels), and $a$ is the acceleration due to gravity (we are equating initial and final kinetic plus potential energy).

Note that in general here there are two reasonable moments to take as the 'final' moment: the moment when the person reaches maximum height (which is useful for the computation in the previous paragraph), and the moment the person is about to again touch the ground. In the former case, the final velocity is zero, while in the latter the final velocity is just opposite the initial velocity (which was non-zero). (I think this is perhaps not clear in the answer by zhutchens1, since they mention two 'final velocities' - one zero and one non-zero). Your final statement is correct to compute the average velocity over the whole jump, basically because by symmetry your velocities on the way back down cancelled those on the way up. But in order to compute the instantaneous final velocity you would need to use the SUVAT equations or similar, as hopefully was clear above.