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**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? Im 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

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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

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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

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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 cant know it. My teacher mentioned this is class but I have no idea on how else to do it :(

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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

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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

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