When to use which kinematic equation?

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Homework Statement

A stone is flung straight up from a point 1.50 m above the ground and with an initial speed of 19.6 m/s.
a. What is the stone's maximum height above the ground?
b. How much time passes before the stone hits the ground?

Homework Equations

The four kinematic equations

The Attempt at a Solution

I used the kinematic equation: x=1/2(v+vo)t, solved for t, and got 2 s. (I used x=21.1, v=0, and vf=19.6). The answer book used x=xo+vot+1/2at^2, and got 4.08 s.

My REAL questions are:
1)why is it better to use the equation they did?
2)shouldn't all of the kinematic equations given the same answer? Does it have to do with the eqn I used had velocity, while theirs had distance?
  • #2
Where did you get x=21.1 from? What are the units on x? What's vf? The equation you cited has v and v0.

The kinematic equations are consistent if you use them correctly.
  • #3
Where did you get x=21.1 from? What are the units on x? What's vf? The equation you cited has v and v0.

The kinematic equations are consistent if you use them correctly.

Here's how I got 21.1. The units are meters. In order to solve for total distance as the problem asks, I used the equation v^2=vo^2+2ax. Solving for x gives: x=(v^2-vo^2)/(2a). I used v=zero (an assumption on my part), vo=19.6, and a=9.8m/s^2. Plugging all of this in gave x=19.6m. I added this value to 1.50 because the problem said it was launched 1.50 metres above the ground; total distance = 21.1 m. I subsequently used this value for d in the second part of the question.
  • #4
OK, I think you're just misinterpreting the question. They're asking you to find the total time from when the stone is released until it hits the ground. You actually calculated the time it takes the stone to go from the apex of its trajectory until it reaches the height from which it was released. Do you see why that is?
  • #5
well, following what you're saying, the v+vo part of the equation I just used corresponds to the "start" and "stop" regions of the ball's path, which is really just saying half the distance of the ball's total path, as it has to travel back down again (notwithsatnding the 1.50metres). I guess I just assumed that because I plugged in 21.1 m (the TOTAL distance), this value would compensate. Maybe I should have just multiplied the calculated value of t by 2?
  • #6
What I said in my previous post was slightly wrong. The time you calculated is only approximately equal to the downward flight time. (I was thrown off by you saying you found t=2.00 s. Using your numbers and formula, you actually get t=2.15 s, not t=2.00 s.) The reason for this is because your numbers aren't consistent with the actual flight of the stone. When the stone's downward speed is 19.6 m/s, the same as its initial upward speed, it's only fallen back to its original height, which is 1.5 m above the ground, so it's actually only fallen 19.6 m, not 21.1 m. So the numbers you're plugging into the equations aren't consistent with the stone's flight.

Physics texts are often kind of loose with the notation, which can unfortunately lead to misconceptions. The equation you used should actually say

[tex]x-x_0 = \frac{1}{2}(v+v_0)t[/tex]

The pairs x, v and x0, v0 need to match up. In other words, when the stone's displacement is x0 or x, its speed is, respectively, v0 or v. In this problem, when the stone has fallen 21.1 m from the apex, it's moving faster than 19.6 m/s, so you can't plug in the pair x=21.1 m and v=19.6 m/s and get the correct answer.

So getting back to your original questions, the main reason the equation you used isn't particularly helpful is that you don't know the final speed v of the stone right when it hits the ground. On the other hand, you do have all the quantities needed to use the equation the book used. Also, your approach only gives you part of the time of flight (if you had the correct final speed v). You'd still have to do another calculation to find the total flight time. There's nothing wrong, per se, with doing it this way, but it's unnecessary to break the trajectory up this way. It usually just makes the calculations more complicated.

Also note that the x that appears in the equations isn't distance; it's the displacement. Similarly, v isn't the speed, but the velocity. These are vector quantities. If you're consistent with the signs, your calculations will work out without resorting to breaking the trajectory up into different pieces.

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