Normal Acceleration of a Thrown Ball: What Is It?

[diredragon]

Homework Statement

a ball is thrown from some height horizontally with some velocity. After 1 second what is its normal acceleration?
Ignore air friction

Homework Equations

The Attempt at a Solution

I was absolutely certain that it is 10m/s2 couse that doesn't change and it has no acceleration in horizontal direction. But somehow the answer is $5\sqrt{2}$ what does this mean?

 

[Doc Al]

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

 

[diredragon]

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

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

 

[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.

 

[Doc Al]

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

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

 

[diredragon]

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

Is this the picture of the problem?( uploaded )
Well in our case, there isn't any tangential acceleration right?
I can't seem to find a way to calculate the normal acceleration using only the 10m/s couse that doesn't change...what can that tell me?

 

[Bandersnatch]

there isn't any tangential acceleration right?

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.

 

[Doc Al]

Well in our case, there isn't any tangential acceleration right?

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

I can't seem to find a way to calculate the normal acceleration using only the 10m/s couse that doesn't change...what can that tell me?

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

 

[diredragon]

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.

"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?

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?

 

[Doc Al]

Isnt the gravitational acceleration the one that curves the path.

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

If i find the angle between $a_n$ and $g$ its simply $gcosx=a_n$ right?

Right!

 

[diredragon]

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

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?

 

[Doc Al]

So it should be 45 degrees as the velocity in the x and y are 10m/s.

Right.

But what if we are taken 1 s more? How does the angle depend on the time?

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

 

[diredragon]

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

I see, $V_y=V_tcosx$

 

[Doc Al]

I see, $V_y=V_tcosx$

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.