Finding the initial velocity of a projectile motion

[midge]

Homework Statement

A baseball is hit at an angle theta. After a time, it lands 7.5m above the point at which it was hit, the land angle is 28 degree below the horizon. The final velocity is 36m/s2. Find the direction and magnitude of the initial velocity.

Know: vf = 36m/s2
theta (final)= 28 degree
H= 7.5m
Unknown: Vo and theta(initial)

Homework Equations

Vxf = Vxi= Vfcos(theta)
Vyf= Vfsin(theta)
Vy^2= [V(initial)sin(theta)]^2 - 2g(yf-yi)

The Attempt at a Solution

I'm breaking down the initial velocity to its component. First I'm finding the x-component of the initial velocity, which is the same as the final Vx since there is not acceleration in the x-direction:

Vx (initial) = Vfcos(theta)
= 36m/s2 * cos(28)
= 31.79 m/s2

Then I'm looking for the y-component of the final velocity, which is:
Vy= Vfsin (theta)
= 36m/s2 *sin(28)
= 16.9m/s2

From there, I find the y-component of the initial velocity:

Vy^2= [Visin(theta)]^2 - 2g(yf-yi)
16m/s2 = Vy(initial)^2 - 2g(7.5)
Vy(initial)^2 = 432.61
Vy(initial)= 20.8 m/s2

To find the initial velocity, I used a2+b2 = c2 for the magnitude and tan(theta)= Vi/Vx for the direction, and the final answer I found is 38 m/s2 at the starting angle of 33.19


***My concern is that I based my solution on the assumption that the landing angle is 28 and so I use that for all my final thetas. Also, I'm not sure using the last equation, to find the initial Vy is valid. Please let me know where I went wrong in this approach. Thank you.

 

[The Liberator]

When the baseball is hit at $$\theta$$ degrees to the horizontal, does the baseball reach it's maximum height (7.5 m) and fly down again?

I am also a bit puzzled about what you mean the 28 degrees is.

 

[Mentallic]

You approached the question in the same way that I did. Since the question stated that the landing angle is 28^o at 7.5m height (I assumed it hit a tall ceiling), there is nothing wrong with using it to solve the problem. In fact, it's necessary to be used. Also, other assumptions are made such as g=-9.8ms^-2, no air resistance and thus V_xi=V_xf

The only problem I noticed is that you kept saying $v=x.m/s^2$. Velocity is a speed, not acceleration. $v=x.m/s$ would be correct.

 

[The Liberator]

Yeah, as said, I think the question needs to be clarified somewhat, as I am slightly confused to how you came to get the final velocity and its' components.

 

[midge]

Liberator,

The 7.5m is not given as the maximum height. It is how far above y=0 it landed. The problem does not state what the maximum height it, it just say that y=7.5 at the landing point. I am uncertain about the 28 degree angle as well. I don't quite understand what "28 degree below the horizon mean," does it mean I use 28 for my angle theta or another number?

 

[midge]

Metallic,

yea...it's whether I need to use 28 or another angle that I'm concerned about. And thank you regarding the m/s for velocity. I keep missing that =)

 

[The Liberator]

I assume it means that the baseball landed at to the horizontal 28 degrees. Also, know that the 7.5 m is how much higher the ball landed at the end will make it a lot harder.

 

[Mentallic]

The question is fine the way it is laid out; you just need to get your head around it.

Think of this: A person hits the baseball at angle $\theta$ with an unknown velocity (but let's assume it's large). Now let's also assume there is a ceiling that is 7.5m above the point where the ball was hit. When the ball hits the ceiling, there is an angle of 28^o made with the ceiling and the direction of the ball at the point of contact. The ball hits the ceiling (at that fairly sharp angle) with a velocity 36m/s.

Thats pretty much the question, but reworded. More clear now?

 

[midge]

yes, thank you