# How to find the time for a ball with (v_i=/=0) to drop?

1. Mar 6, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult. Suppose a ball is spiked from a height of 2.30 m with an initial speed of 20.0 m/s at a downward angle of 18.00°. How much farther on the opposite floor would it have landed if the downward angle were, instead, 8.00°?"

$y_0=2.3m$
$y=0m$
$x_0=0m$
$v_0=20\frac{m}{s}$
$θ=18°$

2. Relevant equations
$x-x_0=v_0t+\frac{1}{2}at^2$

$\vec s_0=(2.3m)j$
$\vec s=(s_x)i+(0m)j$
$\vec v_0=(19.02\frac{m}{s})i+(-6.18\frac{m}{s})j$
$\vec a=(9.8\frac{m}{s^2})j$

3. The attempt at a solution
So I've basically separated each of the positions, the initial velocity, and acceleration into four vectors. Since horizontal motion does not at all, affect vertical motion, I think I can use the usual SUVAT formulas to find the time and the distance $s_x$ using vectors. Since a ball with an initial vertical velocity of $-6.18\frac{m}{s}$ but no horizontal motion would have the same final velocity as a ball with the same vertical velocity plus horizontal motion. Yet, when I plug the numbers in, I don't quite get a sensible answer. Using the equations for only the y-parts of the factors involved...

$0m-2.3m=(-6.18\frac{m}{s})t+\frac{1}{2}(9.8\frac{m}{s^2})t^2$
$-2.3m=(-6.18\frac{m}{s})t+(4.9\frac{m}{s^2})t^2$
$4.9t^2-6.18t+2.3=0$

When I use the quadratic equation, it doesn't work out. The only way it could work out is if acceleration were negative, which I doubt.

$t=\frac{6.18±\sqrt{38.2-45.08}}{2(4.9}$

So yeah; imaginary numbers and all that. Could someone tell me what I'm doing wrong?

2. Mar 7, 2016

### RUber

It looks like you have the wrong sign on your velocity...or your gravity...they should be going in the same direction, right?

3. Mar 7, 2016

### Eclair_de_XII

Oh, of course; that makes sense.

But if acceleration is negative, then doesn't that mean that the speed is decreasing? It's increasing, which is why I wrote acceleration as positive.

4. Mar 7, 2016

### Eclair_de_XII

$-4.9t_1^2-6.18t_1+2.3=0$
$t_1^2+1.26t_1-0.47=0$
$t_1=\frac{-1.26±1.86}{2}=0.3s$
$x_1=(20\frac{m}{s})(cos(-18°))(0.3s)=5.717m$

$-4.9t_2^2-2.78t_2+2.3=0$
$t_2^2+0.57t_2-0.47=0$
$t_1=\frac{-0.57±1.48}{2}=0.458s$
$x_2=(20\frac{m}{s})(cos(-8°))(0.458s)=9.064m$

$Δx=9.064m-5.717m=3.347m$

I still don't understand why acceleration would also be negative if the ball is gaining in speed.

5. Mar 7, 2016

### RUber

The velocity is increasingly negative. That happens when you have a negative velocity and a negative acceleration.
If the signs of v and a are the same, speed is increasing. If they are different, then speed will decrease.

6. Mar 7, 2016

### Eclair_de_XII

That is a very useful thing to know. I must be sure to write it down somewhere.

7. Mar 7, 2016

### BvU

That's easy to write down: $$\vec a = {d\vec v\over dt}$$ or, in case calculus is still ahead of you :$$\vec a = \lim_{\Delta t\downarrow 0} {\Delta \vec v\over \Delta t}$$

8. Mar 7, 2016

### Eclair_de_XII

Yeah, I guess the negative sign would carry over after differentiation, since t isn't to a negative power.

I'm in Calculus II right now, by the way.

9. Mar 7, 2016

### Eclair_de_XII

Hold on a sec; what if the ball is decelerating, and the velocity is positive and the acceleration is negative? How would the latter be a derivative of the former if they haven't got the same signs?

10. Mar 7, 2016

### SteamKing

Staff Emeritus
Since when does the derivative of a function have to be the same sign as the original function?

Draw a plot of velocity versus time, where the velocity starts out at a relatively high positive number and then decreases as time increases. Draw a tangent to the velocity curve where it is decreasing but all of its values are positive. What does the slope of the tangent line represent? Is the slope positive or negative?

11. Mar 7, 2016

### Eclair_de_XII

Oh, of course; how silly of me. I already knew this, but didn't really apply it. The slope of the tangent line represents the derivative of the function, and just because the original function is greater than zero, doesn't mean it's not decreasing. Which is to say, it doesn't mean that the derivative is positive. I can't help but think, though, that at some time intervals, acceleration is negative, while at other intervals, it's positive. Like when you throw a ball into the air. Its acceleration is negative, then zero at maximum height, then positive when it comes down. So the question I'm asking, is how would you determine, for a certain movement, whether acceleration is positive or negative, if it changes sign in the midst of its action? I'm going to guess that in that ball example, I'd need to separate it into two separate equations?

12. Mar 7, 2016

### SteamKing

Staff Emeritus
This is why applying formulas to physics problems blindly is a bad idea. You should think about the physical processes which are occurring before you dig into your bag of formulas and other mathematical tools.

To answer this particular question about tossing the ball, yes, it's better to analyze the ball going up separately from the case of the ball coming down.

13. Mar 8, 2016

### RUber

Don't confuse acceleration and velocity. For problems like tossing a ball, you have a positive initial velocity subject to a constant negative acceleration from gravity.
At the top of the path, the ball has lost all of its initial velocity. The same negative acceleration is still working on the ball and causes it to start moving in the negative direction.
$v(t)=v(0)+at$