How Does Throwing a Ball from a Moving Truck Affect Its Velocity to an Observer?

[CursedAntagonis]

Homework Statement

A student is riding in the back of a truck traveling along a straight level road at Vt = 20 m/s. He faces toward the rear of the truck and throws a ball with a velocity of 15 m/s at an angle of 37 degrees above the horizontal relative to the truck. He releases the ball 2.0 meters above the ground level. Ignore air resistance.

What is the horizontal velocity of the ball relative to an observer at roadside?

The Attempt at a Solution

I don't even know where to start this.

I am assuming I need to find the balls horizontal velocity and subtract it from the initial 20 m/s since the student threw the ball in the opposite direction. Then figure out the horizontal velocity relative to the observer...

 

[Hootenanny]

I am assuming I need to find the balls horizontal velocity and subtract it from the initial 20 m/s since the student threw the ball in the opposite direction. Then figure out the horizontal velocity relative to the observer...

Sounds good to me, so how would you go about finding the horizontal velocity of the ball relative to the truck?

As an aside, I think it would be a good idea to set up a definite coordinate system. Let's say that the truck is traveling in the positive x direction and the boy throws the ball in the negative x direction.

 

[CursedAntagonis]

Sounds good to me, so how would you go about finding the horizontal velocity of the ball relative to the truck?

As an aside, I think it would be a good idea to set up a definite coordinate system. Let's say that the truck is traveling in the positive x direction and the boy throws the ball in the negative x direction.

Gotcha. The problem provided me with a diagram (I was not aware if I am allowed to simply scan the problem and post it so I figured I should not risk it) The truck is going in the positive X velocity and the ball is thrown in the negative X velocity.

I found the horizontal velocity to be approximately 12 m/s by setting Cos37 = x/15...is that correct?

I was wondering how I could set this up in variables? then plugging in the numbers. My teacher likes us to work out the problems with the variables and finding the answer that way prior to just punching in numbers.

 

[Hootenanny]

Gotcha. The problem provided me with a diagram (I was not aware if I am allowed to simply scan the problem and post it so I figured I should not risk it) The truck is going in the positive X velocity and the ball is thrown in the negative X velocity.

I don't see there being any copyright issues, I'm sure this falls under fair-usage.

I found the horizontal velocity to be approximately 12 m/s by setting Cos37 = x/15...is that correct?

Sounds good to me (although I'd keep a few more sf's for the full calculation).

I was wondering how I could set this up in variables? then plugging in the numbers. My teacher likes us to work out the problems with the variables and finding the answer that way prior to just punching in numbers.

If your finding it difficult to set up an equation, it is often easier to start with a word equation;

v_observer = v_truck + v_{x ball}

Can you go from here?

 

[CursedAntagonis]

Could I set it up as this?

Vbe = velocity of the ball relative to earth
Vte = velocity of the truck relative to earth
Vbt = Velocity of the ball relative to the truck

so I get Vbe = Vte - Vbt which in return gives me the velocity of the ball relative to the observer...

did I get this right or am I way off?

 

[Hootenanny]

Could I set it up as this?

Vbe = velocity of the ball relative to earth
Vte = velocity of the truck relative to earth
Vbt = Velocity of the ball relative to the truck

so I get Vbe = Vte - Vbt which in return gives me the velocity of the ball relative to the observer...

did I get this right or am I way off?

Your spot on except for the sign, note that the ball's velocity relative to the truck will be negative. So now all you need to do is correct the sign and sub in your values.

 

[CursedAntagonis]

Your spot on except for the sign, note that the ball's velocity relative to the truck will be negative. So now all you need to do is correct the sign and sub in your values.

Ok so would it be this?

Vbe = 20 + 15

so the velocity relative to the observer is going to be 35 m/s?

How come it's adding it and not subtracting it? I was assuming it would have to be 5 m/s.

I'm sorry I'm a bit slow with questions like these.

 

[Hootenanny]

Ok so would it be this?

Vbe = 20 + 15

so the velocity relative to the observer is going to be 35 m/s?

How come it's adding it and not subtracting it? I was assuming it would have to be 5 m/s.

I'm sorry I'm a bit slow with questions like these.

Close, but like I said above, your equation should be Vbe = Vte + Vbt with Vbt = -15m/s, which would give your 5m/s as you expected.

 

[CursedAntagonis]

Close, but like I said above, your equation should be Vbe = Vte + Vbt with Vbt = -15m/s, which would give your 5m/s as you expected.

Yes yes, I put that negative in the original equation as I was expecting the 15 to be negative. I guess that's how I ended up confusing you and myself =P

so the horizontal velocity of the ball relative to the observer is 5 m/s? That's what we found I hope lol.

The next question is With what velocity (magnitude and direction) is the ball projected as seen by the roadside observer?

I would have to use cos 37 = 5/v correct? which would give me 6.26 m/s. But the problem is that the ball is 2 m above (since the ball is thrown from the back of the truck)...

 

[Hootenanny]

Yes yes, I put that negative in the original equation as I was expecting the 15 to be negative. I guess that's how I ended up confusing you and myself =P

Yes, but if we used your equation with Vbt = -15m/s we would have;

Vbe = Vte - Vbt = 20 - (-15) = 20+15 = 35m/s

You should write your equations first and then just sub the values in as you have them. If you start messing around with signs in your equation your going to get yourself in trouble.

so the horizontal velocity of the ball relative to the observer is 5 m/s? That's what we found I hope lol.

Sounds good to me

The next question is With what velocity (magnitude and direction) is the ball projected as seen by the roadside observer?

I would have to use cos 37 = 5/v correct? which would give me 6.26 m/s.

Sounds good to me, just don't forget you quote your angle.

But the problem is that the ball is 2 m above (since the ball is thrown from the back of the truck)...

Why would this affect the velocity?