Physics: Projectile Question.

Question

For a projectile shot into the air, at any instant on the way up or down, 's' is the

scalar value of the displacement: it is the distance the object is from the starting

point. It is not the total distance traversed to get there. When a projectile is thrown

straight up, reaches peak altitude, and comes halfway back down, the displacement

1/2 of 's' at that moment is one third of the distance traveled.

"The one third of the distance traveled" of the sentence is really confusing me. Wouldn't the distance be at least 2/3 of the distance traveled with 1/3 remaining? Or, perhaps, what I was thinking it would be was 1 ½ distance traveled with 1/2 distance remaining. Am I missing something?

Answer 1

You are thinking about the total distance traveled compared to the total distance of the trip from start back to ground. This statement refers to the displacement (which is final position minus initial position) compared to the total distance traveled (ground to "midair on the way back down"). The displacement is small compared to the total distance traveled (ground to "midair on the way back down") so 1/3 fits.

Mathematically:

For an object thrown straight up, we can define it's height by:

h(t) = h₀ + v₀t + 1/2at² where "t" is the time the object is in the air, h0 is the initial height (which we can consider as zero if our coordinate plane is placed at the location of "launch", v0 is the initial velocity, and a is the acceleration due to gravity (aka "g")

So h(t) = v₀t + 1/2gt²

Considering that the object must travel up to the vertex then back down to the halfway point, we can find the vertex by using t = -b/(2a) = -v₀/(2(1/2g)) = -v₀/g. To find the height, plug in t = -v₀/g to get:

h = v₀(-v₀/g) + 1/2g(-v₀/g)² = -v₀²/g + v₀²/(2g) = -2v₀²/(2g) + v₀²/(2g) = -v₀²/(2g) [Note that g is a negative number so this result will be positive)

So the object travels a distance of "-v₀²/(2g)" from the ground to the top of it's flight. The distance halfway back is [-v₀²/(2g)]/2 = -v₀²/(4g) so the total distance traveled is -v₀²/(2g) + -v₀²/(4g) = -3v₀²/(4g)

The displacement is the final position - initial position = -v₀²/(4g) - 0 = -v₀²/(4g)

So the displacement to total distance traveled is (-v₀²/(4g))/(-3v₀²/(4g)) = 1/3