Distance ball travels question

1. Sep 17, 2006

phantomcow2

I have a homework problem I was told to "think about". The teacher says he did not have time to teach us a few things we should know about this, but we should think about it over the weekend so we can do it on monday. Well I am trying, but I feel like there is information I am missing. Not information in the problem, but a method to solve.

A ball is given an initial velocity of 34m/s at an angle of 30degrees. Determine the distance the ball travels before it hits the ground.

Can somebody give some steps of how I would best solve this problem? We just started two dimensional motion, with your I and J vectors.

2. Sep 17, 2006

Hootenanny

Staff Emeritus
The information given is sufficient to solve the question. Do will need to split the initial velocity into two components, one vertical and one horizontal. Once you have done this you should use basic kinematics to produce two equations of motion (one for each component), once you have done this you should be able to solve the vertical component for the time of flight t and the horizontal component will then give you the distance travelled.

If you wish to attempt the question here, we will be happy to guide you through.

3. Sep 17, 2006

phantomcow2

Okay so using trig, drawing a right triangle of vectors, I got a few velocities.
We know that the ball is thrown at 30 degrees, speed of 34m/s.
So Sine(30 * 34

17m/s as a velocity of our vertical.
29.45m/s as a velocity of our horizontal.

I would like to attempt it here if possible. THe equation I was thinking of was
Distance = Initial distance + Initial velocity * time + 1/2*a*t^2

But clearly that won't work because I have distance and time as a variable. So can you tell me what equation it is I would need to use? This may be one I have not yet encountered

4. Sep 17, 2006

Hootenanny

Staff Emeritus
Okay, lets look at each component independently. I am assuming here that the ball was launched from the ground and not an elevated position.

Vertically

As the ball is launched from the ground and returns to the ground, you know that the displacement is 0m, yes? You also know the initial velocity that you correctly calculated. You also know that there is a constant acceleration (due to gravity). So the equation you stated;
Is correct, but note that you should take the displacement and not the distance. Now you have only one known allowing you to solve for flight time.

Horizontally

Now, horizontally we have the initial velocity and we need to find the final displacement. Since there is no acceleration in the horizontal plane (ignoring air resistance) the ball will travel at a constant velocity, allowing use easily to calculate the distance travelled.

Do you follow?

5. Sep 17, 2006

phantomcow2

You are correct to assume that the ball is launched from a ground position. So since there is no gravity in our horizontal, obviously our equation is simplified.

So you can use the displacement for the distance. But, do I know the displacement? Wouldn't I need to know the end point of where the ball would land and then use the distance formula to determine displacement?

Thanks for the help

6. Sep 17, 2006

Hootenanny

Staff Emeritus
Not if you just consider the vertical plane. In the vertical plane the ball started at ground level and it will obviously finish at ground level when it impacts. Since vertically it landed at the same level it started it's vertical displacement is zero. Do you follow?

7. Sep 17, 2006

phantomcow2

Ah! That makes sense now.

So I can plug in 0 for my Distance for hte vertical equation. ANd now I have the perfect quadratic with a small bit of rearranging. I can solve for T.
NOw with T, I can plug this into my horizontal equation, correct?

8. Sep 17, 2006

Hootenanny

Staff Emeritus
Sounds spot on to me