Help! Physics Problem has me Stumped!

[phantomcow2]

I have to say, Physics has been going well for me. But he assigned a problem in preparation for a test. Well needless to say, I am not sure where to begin! I was hoping somebody could at least point me in the right direction.

A baseball is hit for a home run, it travels 138m. It just clears a 6.5m wall when it lands, where it is caught by a lucky fan. So the ball essentially starts off at point (0,0), and ends at (138,6.5).
It is hit at a 40 degree angle. Calculate initial velocity and total flight time.

I know it must be possible. Because only at one velocity at 40degrees will the ball actually go 138m. Like if it was hit at 1m/s, obviously it would not go the full 138. It's finding this balance point. My thought was to first find velocity, THEN find time. If only i knew how

I've got a few ideas brewing upstairs, but ugh.

 

[radou]

Step 1: present your work, and we will be glad to help you. (Hint: write down the equations of velocity and position of the ball.)

 

[phantomcow2]

I am stuck because, well I want to break it down into two vectors. So say I use X1=X0+V0t+1/2at^2

For one vector, the j vector because it has acceleration. I don't know time or initial velocity, I don't understand how I find what I need to find.
So 6.5=0 + Vot + -4.905t^2

I would think I need initial velocity to calculate this.

 

[phantomcow2]

Okay to show prior efforts..
Here is a link to a thread I started in another forum I participate in. Same problem. http://www.bikeforums.net/showthread.php?t=231977

 

[radou]

Well, you just have to solve a system of two equations with two unknowns, which are the initial velocity and the total 'flight time'. The equation of motion for the x-direction is x(t) = v0*cosA*t , and for the y-position y(t) = v0*sinA*t - 1/2*g*t^2. (A is, of course, the angle of the initial velocity.) Now, since you know the x and y coordinate at the end of the 'flight', just plug them in and solve the system of equations.

 

[andrevdh]

The parabolic equation for a projectile is of the form

y = Ax - Bx^2

where

A = \tan(\theta _o)

and

B = \frac{g}{2(v_o \cos(\theta _o))^2}

Another solution:
Shoot the monkey - d is the distance the poor creature drops through before encountering the projectile. h the height at which the monkey is shot: 6.5 meter. Then

\tan(\theta _o) = \frac{d+h}{138}

the rest is easy ... or is it? I do get the feeling that this is the appoach your teacher expects from you.