# Displacement and position for a ball thrown up into the air

What would the displacement and distance traveled (from the starting point to the ending point) be for a ball that is thrown up into the air? (Hint: Think about the definition for displacement)

I can't wrap my head around this problem. I know that the displacement is the shortest distance between two points, and the distance traveled is the actual distance. Thus, the displacement would have to be $0$ meters since a ball thrown into the air would come down and the shortest distance is $0$.

The definition for displacement can also be represented by $s = v_i(t_f-t_i) + 0.5a(t_f-t_i)^2$. In that case, I don't see how $s$ could equal $0$ because the change in time, initial velocity, and acceleration (force of gravity) all need to be finite numbers. This is where I get confused, because does $s$ represent the displacement or the total distance?

Also, if we were calculating the average velocity, which is displacement over time, then if displacement was zero, wouldn't the average velocity also be zero, which also doesn't make sense?

• Note that $v_i$ and $a$ might have opposite sign. – garyp Jun 5 '17 at 0:42
• What would $a$ be, $+9.81 m/s/s$ or $-9.81 m/s/s$? – Stardust Jun 5 '17 at 0:44
• Depends on whether $v_i$ is positive or negative. – Bill N Jun 5 '17 at 1:14
• The sign of $g$ depends on how you set up your coordinates. If up is positive, then $g$ is negative. If up is negative, then $g$ is positive. – garyp Jun 5 '17 at 2:06

$$\frac{d}{dt}(\frac{1}{2}at^2+v_0t) = 0$$
You can solve for $t$ to find the time at which the ball reaches its max height. Plugging it back in to the original equation will reveal the actual maximum height, which you can double to get the total distance traveled.