# Normal acceleration

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1. Jun 21, 2016

### diredragon

1. The problem statement, all variables and given/known data
a ball is thrown from some height horizontally with some velocity. After 1 second what is its normal acceleration?
Ignore air friction
2. Relevant equations

3. The attempt at a solution
I was absolutely certain that it is 10m/s2 couse that doesnt change and it has no acceleration in horizontal direction. But somehow the answer is $5\sqrt{2}$ what does this mean?

2. Jun 21, 2016

### Staff: Mentor

They probably want the component of the ball's acceleration normal to its path at that point. (What was the horizontal velocity?)

3. Jun 21, 2016

### diredragon

It was 10m/s so how is that done?

4. Jun 21, 2016

### Bandersnatch

Start by drawing a free body diagram after 1 second. Mark the acceleration and its tangential and normal components. See if you can find a way to calculate $a_N$ using geometry.
You will have to find a way to ascertain the angles between various acceleration components. What you know about velocity should help.

5. Jun 21, 2016

### Staff: Mentor

Figure out the direction of the velocity vector at that point.

6. Jun 21, 2016

### diredragon

Is this the picture of the problem?( uploaded )
Well in our case, there isnt any tangential acceleration right?
I cant seem to find a way to calculate the normal acceleration using only the 10m/s couse that doesnt change...what can that tell me?

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7. Jun 21, 2016

### Bandersnatch

There is. It's the component of gravitational acceleration along the velocity vector. It's what makes the object accelerate as it follows the curved path. Normal acceleration curves the path.

8. Jun 21, 2016

### Staff: Mentor

"Tangential" means tangential to the path: Parallel to the velocity vector at that point.

What direction is the normal to the velocity vector at that point? What angle does that normal make with the direction of gravity?

9. Jun 22, 2016

### diredragon

Isnt the gravitational acceleration the one that curves the path.
If i find the angle between $a_n$ and $g$ its simply $gcosx=a_n$ right?

10. Jun 22, 2016

### Staff: Mentor

Yes, it's the gravitational force acting normal to the velocity that curves the trajectory.

Right!

11. Jun 22, 2016

### diredragon

So it should be 45 degrees as the velocity in the x and y are 10m/s. But what if we are taken 1 s more? How does the angle depend on the time?

12. Jun 22, 2016

### Staff: Mentor

Right.

You tell me. Only one of the velocity components will change.

13. Jun 22, 2016

### diredragon

I see, $V_y=V_tcosx$

14. Jun 22, 2016

### Staff: Mentor

OK. Here's what I would do, for any time $t$:
$V_x$ is constant
$V_y = gt$ (downward)

Then use a bit of trig to find the angle from the vertical.