Check my Throw the Ball lab. (2D kinematics, force, work)

[ZamielTheGrea]

Check my "Throw the Ball" lab. (2D kinematics, force, work)

Here I have a lab, with explanation, questions and my worked out answers (tried to make it as neat as possible with all the formulas I use, lowercase letters are subscript letters). Can someone carefully check my math for all this? I am especially worried that my vector angle in #3 was incorrectly calc'd...
So, we went outside, and had a partner throw a baseball. The other recorded time from release to contact with the ground by the ball. Distance the ball flew is also recorded. Afterwards, several other measures are taken and everything is listed below:

T2 = 1.34s (time the ball was in the air)
X = 19.3m (distance horizontally traveled by the ball)
Y = 1.88m (vertical height of the ball's release point)
D = 0.4m (distance the ball travels while in the hand)
ΔH = 0.34m (vertical distance the ball travels while in the hand)
M = 140.5g (mass of baseball)

g = -9.8
no air resistance
------------------------------------------------------
Calculations:

1) Determine the horizontal component of the velocity, Vx, of the ball after release.
X = VixT
19.3 = 1.34Vx
Vx = 14.4 m/s

----------------------------

2) Determine the vertical component of the velocity, Vy, of the ball after release.
Y = ViyT + 0.5AYT^2
-1.88 = 1.34Viy + 0.5(-9.8)(1.34^2)
-1.88 = 1.34Viy - 8.8
Viy = 5.16 m/s

------------------------------

3) Determine the size of and direction of the velocity, V, of the ball after release.
V = sqrt(14.4^2 + 5.16^2)
V = 15.3 m/s

sinΘ = opp/hyp
sinΘ = 0.34/0.4 = 0.85
Θ = 58.2° up right
^ Problem that may be here is that I calculate angle BEFORE release:

```````/-|
``````/--|
``0.4`/---|
````/-----| 0.34
``/-------|
`/--------|
/-Θ-------|
-----------
(triangle o.o)

The hypotenuse in real life is an arc, and the angle changes continuously, so upon release I really can't know it. My teacher mentioned this is class but I have no idea on how else to do it :(------------------------------

4) Use the open system version of the conservation of energy equation to determine the force, F, with which the ball was thrown.
KE = 0.5MGH
KE = 0.5(0.14)(15.3)^2
KE = 16.4 J

PEg = MGH
PEg = 0.1405(9.8)(0.34)
PEg = 0.468 J

ME = KE + PE
ME = 16.4 + 0.468
ME = 14.868 J

W = FDcosΘ
14.868 = F0.4 cos58.2°
F = 70.54 Newtons

------------------------------

5) Calculate the time, T1, the ball is in the hand being thrown.
F = MA
70.54 = 0.1405A
A = 502.064 m/s^2

Vf = Vi + AT
15.3 = 0 + 502.064
T = 0.0509s

 

[BruceW]

I think you've got questions #1 and #2 correct. The method you have used in #3 might be a good approximation if the ball travels in a roughly straight line while the person is throwing it. But as your teacher said, it will generally be a curved path. There is another way to calculate #3 (hint - use the answers you got in questions #1 and #2).

In question #4, On the first line of calculation, I think you meant to write 0.5MV^2 (not 0.5MGH). But you did the working correctly, so I think you just made a typing mistake? Anyway, I think your answer of ME = 14.868 J (the work done by the guy) is correct. But I don't think your calculation of the force is correct. work is force times distance, right? so why is it multiplied by the cosine?

For question #5, I think you have used the equations correctly. But I think your value for the force was incorrect, so your answer for #5 will have come out wrong. Also, I think there is a more straightforward way of doing this question, but your way should work.