# When to use which kinematic equation?

6FX

## 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?

Staff Emeritus
Homework Helper
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.

6FX
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.

Staff Emeritus
Homework Helper
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?

6FX
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?

Staff Emeritus
Homework Helper
$$x-x_0 = \frac{1}{2}(v+v_0)t$$