Wednesday

August 20, 2014

August 20, 2014

Posted by **Amanda** on Tuesday, November 12, 2013 at 7:40pm.

The data required for this lab is the horizontal displacement of the ball from the location where it was hit to the point where it first strikes the ground and the time that the ball spends in the air. You may assume that the ball was hit from an initial height of 1.5m

t = 3.60 s

△X = 48.9 m

Analysis:

Mathermatical:

*Determine the initial velocity of the ball as it left the bat.

*Determine the maximum height reached by the ball.

*Determine the horizontal distance at which the maximum height was achieved.

*Divide the time that the ball spent in the air into 10 equal interval.

*Determine the displacement, velocity, and acceleration of the ball in both the X and Y directions for each of these time intervals.

*Display the information above in a table.

*Show one set of sample calculations.

Graphical:

*Contrast (x vs t),(v vs t), and (a vs t) graphs for the ball in both the X and Y directions.

*Describe the shape of each graph and explain why each graph has the shape that it does.

Conclusion:

Write summary of your findings from this exercise.

Please Help, answer.

Thank You

- Fly Ball Lab - Projectile Motion Please HELP FAST! -
**Henry**, Thursday, November 14, 2013 at 4:50pmDx = Xo * T = 48.9 m.

Xo * 3.60 = 48.9

Xo = 13.58 m/s = Hor. component of initial velocity.

Tr = Tf = T/2 = 3.60/2 = 1.80 s. = Rise

and fall times.

Y = Yo + g*Tr = 0 @ hmax.

Yo - 9.8*1.8 = 0

Yo = 17.64 m/s = Ver. component of initial velocity.

tan A = Yo/Xo = 17.64/13.58 = 1.29897

A = 52.4o

*Vo = Yo/sin A = 17.64/sin52.4 =

22.3m/s[52.4o]

h max = (Y^2-Yo^2)/2g

*h max = (0-17.64^2)/-19.6 = 15.88 m.

*Dx = Xo * Tr = 13.58m/s * 1.8s=24.44 m.

Use the following time intervals for graphing:

T/10 = 3.60/10 = 0.36 s.

*0.36,0.72,1.08,1.44,1.80,2.16,2.52,2.88,3.24,3.60 s.

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