# Help with a two-dimensional motion problem

1. Jan 8, 2007

### mikesown

1. The problem statement, all variables and given/known data
"A ball player hits a home run, and the baseball just clears a wall 21.0m high located 130.0m from home plate. The ball is hit at an angle of 35.0degrees to the horizontal, and air reisistance is negligable. Assume the ball is hit at a height of 1.0m above the ground

a. What is the speed of the ball?
b. How much time does it take for the ball to reach the wall?
c. Find the velocity components and the speed of the ball when it reaches the wall."

Variables(this was my attempt at interpreting the question):
Delta y = 20.0m
Delta x = 130.0m
theta = 35degrees
2. Relevant equations
This is where I'm confused. The only equation I could come up with was:
$$\Delta y=-\frac{1}{2}g(\Delta t)^2$$

3. The attempt at a solution
The only thing I could do was plug in the numbers(20 for delta y, -9.81 for g),to get $$20.0m=-\frac{1}{2}\left(-9.81\frac{m}{s^2}\right)(\Delta t)^2$$ and solve for t. I got 2.02s, which is not the right answer for part b, which is 3.81s. I would like to understand how to solve the equation, perferably myself, so if someone could give me guidance, rather than a solution, I would appriciate it.

Thanks,
mike

Last edited: Jan 8, 2007
2. Jan 8, 2007

### pmp!

For a):
Could you write the trajectory equation $$y(x)$$, y as a function of x?

3. Jan 8, 2007

### mikesown

To be completly honest, no. My physics class(honors physics) didn't really cover interpreting functions. We worked purely from predetermined equations from the book.

4. Jan 8, 2007

### Littlepig

so, by motion equation: $$x=x_i+v_{i}.t+\frac{1}{2}a.t^2$$

as air resistance is negligable, a is constant, in x and y.
with $$a_y=-9.8$$;

we have:
for x:
$$x=0+cos(35º).v_{i}.t+\frac{1}{2}0.t^2$$ (1)
$$y=1+sen(35º)v_{i}.t+\frac{1}{2}-9.8.t^2$$ (2)

now, we have this:

what speed you want to know? instantaneus speed? speed at exit? speed when ball passes the wall??

you want to know what's the value of "t", when the ball passes at x=130 and y=21

and you have 2 equations with 2 ecognits: t and v_i , you take t and solve b., you take v_i and you solve a.(asuming the inital speed what they are asking, otherwise you must say what it is.. )

by motion equations, you get t and v_i, so, by:

$$v=v_i+a.t$$ you get v(as a is known on both x and y) then by trignometry, you get x and y components...

hope i helped you...

5. Jan 8, 2007

### mikesown

I need to know the inital speed of the ball. Also what is sen?

Last edited: Jan 8, 2007
6. Jan 8, 2007

### cristo

Staff Emeritus
I'll give you a little background, since you say you want to learn and not be told the answers!

The equation that Littlepig gave you is one of the kinematic equations for constant acceleration. These are the equations to use when facing any question on projectile motion like this one. Ill link to them, since it's easier than typing! http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html

When you tackle this sort of problem, you should first write down the horizontal and vertical variables.. both those you know, and those you don't. So, you would write something like vi=.., vf=.., a=.., t=.., d=.. once for vertical and once for horizontal. You should also draw a diagram of the situation, and put on it what the variables correspond to, so you know exactly what you are calculating.

Then, you use the kinematic equations. Look for one that involves the variable you want to find out, as well as other ones you know, as this will be the simplest approach. If this is not available, then you will have to combine equations. However, using these equations properly, you will always be able to solve the problem.

I'll leave the quantitive help to Littlepig, since he has already started.

7. Jan 8, 2007

### mikesown

Ah, I think I see... I know the equation d=vit+1/2at^2.....
Would these two equations work:
$$y=\sin35\,^{\circ}v_{i}t+\frac{1}{2}*9.81\frac{m}{s^2}(t)^2+1$$
$$x=\cos35\,^{\circ}v_{i}t$$

8. Jan 8, 2007

### cristo

Staff Emeritus
Very close. One thing that I didn't mention in my previous post, was making sure you know your which way on the axes is positive and which is negative.

Here, I presume you are taking, in the y direction, upwards to be positive? If so (and I would suggest you do!) then a=-9.81 (acceleration due to gravity acts downwards) and so you equation would be
$$y=\sin35\,^{\circ}v_{i}t-\frac{1}{2}*9.81\frac{m}{s^2}(t)^2+1$$
This one's correct!

9. Jan 8, 2007

### mikesown

Thanks for your help with the equations. I know that a is supposed to be -9.81 but just goofed up ;). How would I go about finding the inital speed of the ball though? Do I set 0=cos35vit? If so, then I still have 2 variables, vi and t. How do I get around this?

10. Jan 8, 2007

### mikesown

I tried to solve for t and got the following:
$$\frac{x}{\cos35\,^{\circ}v_{i}}$$
and
$$\frac{y}{\sin35 v_{i}+\frac{1}{2}*-9.81t}-1}$$
What do I do now?