Baseball's Initial Velocity Problem

[unholymist]

Homework Statement

A baseball is hit at a 45 degree angle 1.3m above the ground. It barely clears a 3m wall 130m away. With what velocity was the ball hit?

Homework Equations

Equations of Motion

The Attempt at a Solution

So I realized that without air resistance this would be a parabolic motion. Since the starting angle is 45 degrees, the 1.7m height (3-1.3) at 130m away cannot be the vertex. I don't think gravity would diminish the vertical force that much. I did some calculations to verify and found the time would be 0.52s for this to be true, making the horizontal velocity 220m/s... making the initial vertical and horizontal not equal... etc.

Anyway I can't see any other way of solving this. I've determined the initial vertical velocity to be Fsin45 and the horizontal to be Fcos45 (F being the magnitude of the 45 degree vector). I can't find any way to isolate F... help?

Thanks. >.<

 

[tiny-tim]

hi unholymist!

(i don't understand how you got the time )

you know the initial velocity in each direction is F/√2 …

so call the time t, do x and y equations, and eliminate t to find F …

what do you get? ​

(btw, why F? most people use V)

 

[unholymist]

hi unholymist!

(i don't understand how you got the time )

you know the initial velocity in each direction is F/√2 …

so call the time t, do x and y equations, and eliminate t to find F …

what do you get? ​

(btw, why F? most people use V)

Well... I know that there's that special triangle with the 45 degree angles but the value wouldn't be exactly √2 here... how would I determine the actual value?

 

[tiny-tim]

Well... I know that there's that special triangle with the 45 degree angles but the value wouldn't be exactly √2 here... how would I determine the actual value?

i don't understand …

cos45° = sin45° = 1/√2, so put time = t and force = F

 

[rwisz]

I would approach this problem by using the equations of motion to analyze the motion of the baseball. In this case, we have the initial angle, height, and distance traveled, and we are looking for the initial velocity.

First, we can use the equation for horizontal distance, x = x0 + v0x*t, to find the initial horizontal velocity (v0x). Plugging in the given values, we get:

130m = 0 + v0x * t

Next, we can use the equation for vertical displacement, y = y0 + v0y*t + 1/2*g*t^2, to find the initial vertical velocity (v0y). Again, plugging in the given values, we get:

1.7m = 1.3m + v0y * t - 1/2*9.8m/s^2 * t^2

Using the fact that the ball barely clears the 3m wall, we can set the final vertical displacement to be 3m. This will give us a second equation:

3m = 1.3m + v0y * t - 1/2*9.8m/s^2 * t^2

Now we have two equations with two unknowns (v0x and v0y) and we can solve for them simultaneously. Solving for v0x in the first equation and substituting into the second equation, we get:

3m = 1.3m + (130m/t) * t - 1/2*9.8m/s^2 * t^2

Solving for t, we get t = 2.64 seconds. Plugging this back into the first equation, we get v0x = 49.24 m/s.

Next, we can use this value of v0x to solve for v0y in the second equation:

3m = 1.3m + v0y * 2.64s - 1/2*9.8m/s^2 * (2.64s)^2

Solving for v0y, we get v0y = 19.6 m/s.

Therefore, the initial velocity of the baseball is 49.24 m/s at an angle of 45 degrees. This means that the initial velocity vector has a magnitude of 49.24 m/s and is directed