As always - not homework. Just trying to get my head around basic math.
OK, so you have an equation that models height of a projectile as a function of time. Let’s say:
h(t)=-5t^2+40t+10 (h in meters; t in seconds)
h’(t)=-10t+40
h’’(t)=-10
So I get that the coefficient in front of t^2 in h(t) is half the acceleration downward due to gravity (to the nearest integer). And the constant, 10, is the height of the projectile at the start of its flight.
But I just realized that the 40 in h(t) is the initial velocity (because h’(0) = 40m/s). Cool. Well, now that I know a smidge about vectors, I should say its initial speed is 40 m/s, right?
Here’s the question: in h(t), what direction is the projectile traveling 40 m/s in? I’m guessing it must be the sum of the horizontal and vertical vectors at the onset of the flight so that h(t) could model an infinite number of combinations of angles the projectile is launched in and horizontal/vertical velocities. So I’m guessing 40^2=(horizontal velocity^2+vertical velocity^2).
The height is not affected by the horizontal speed, only the vertical speed. You have a scalar equation that one presumes is all in the z direction of an x,y,z coordinate system.
Yes, you specified gravity and I assumed we’re using one dimensional vectors, using only the z axis as John Mace stated. Your initial speed will be +40 m/s starting at +10 m. Your acceleration is -10 m/s[sup]2[/sup] so your object will initially rise and slow down, stop, then fall back down.
If you want to use a two dimensional vector, use h(t) = (-5i + xj)t[sup]2[/sup] + (40i + yj)t + (10i + zj) … where x, y and z are horizontal coefficents and i and j are your unit vectors.
You know nothing about the horizontal component of the projectile’s motion, from this equation, and hence know nothing about the launch angle. All you know is the vertical component, and hence what you know about the velocity is its vertical component.
So the 40 m/s is the projectile’s upward velocity. But is that independent of the horizontal velocity or does the horizontal velocity have to be 0?
So could this equation model both of the following scenarios (on an Earth-mass planet with no atmosphere)?
a) a projectile shot straight up from a height of 10m
b) a projectile shot from a height of 10m on a 45 degree angle at 56.6 m/s (which if I’m doing the math correct likewise would have a vertical velocity of 40 m/s as well as a horizontal velocity of 40 m/s).
Thank you! I didn’t appreciate how rudimentary all these examples were in the high school level math I’ve been working on. I don’t know what unit vectors are so I’ll revisit this when I get that far.
You might find this information about unit vectors in physics, actually. I sure don’t remember vector functions coming up in math until I was in college. They starting showing in physics first, answering questions exactly like yours.
You’ve almost answered your own question here. Yes, the vertical velocity and horizontal velocity are entirely independent of one another. And no, the horizontal velocity doesn’t have to be 0; as watchwolf49 wrote, that would only be 0 if the motion is straight up-and-down.
Well, not exactly. The initial speed is however fast it is going in whatever direction it is going when you first launch it. That is, the resultant of the vertical speed and the horizontal speed. But the h(t) function tells you nothing about the horizontal speed, and thus nothing about the resultant. The initial speed of 40m/s is the vertical component only.
yes, a. no horizontal movement
b. it stays at 40m/s horizontally from start to finish, and your formula give you the vertical… position, horizontal position is rather simple… 40t.
My high school (non-calculus-based) physics class definitely used vectors, but I don’t remember seeing vector-valued functions there. Vector-valued functions (including their derivatives and their use to represent things like position and velocity) are typically covered in the third semester of the calculus sequence.
The community college I went to did a variation on that curriculum: All the standard Calculus textbooks had all the vector stuff up in the third-semester area. But there was nothing there that inherently had to wait until then – all that stuff could be covered much earlier.
So at my college, they included those vector chapters in the first semester, specifically for the benefit of all the physics students who needed that stuff right away. I believe it was the physics professors that had pushed for this.
Now, as it happened, I had taken first semester calculus at a different school in a different city in a different state, and I hadn’t learned all the vector stuff. When I enrolled here into second-semester calculus, I had to spend the two weeks before the semester started studying that myself.
My sense is that the student of mathematics would not, generally, have very much difficulty understanding vector functions once the principles in scalar functions are established. The only time vector functions get squirrely is when we try to integrate them, thus third term calculus.