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BitterSuites

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**[SOLVED] 2D Kinematics Help**

Basically I want to make sure that I have worked this correctly and find out what I did wrong in Part 3, as my solution is not logical.

**Problem**Consider a ball thrown up from the ground. It passes a window in the time interval .251s. The distance across the window is 1.49m.

1) Find the average speed as the ball passes the window.

2) What is the magnitude of the decrease of the velocity across the window?

3) If the ball continues its path upward without obstruction, find the travel time between the top of the window and the ball's maximum height.

**My Work**

1)V = (Vo + Vf)/2 = (Xf + Xi)/t (This correspondence was the only thing I could come up with to solve this problem)

5.936 = (Vo + Vf)/2

2(5.936) - Vf = Vo

Vf = Vo +at

Vf = 2(5.936) - Vf + at

2Vf = 2(5.936) + at

Vf = 5.936 + (at/2)

Vf = 5.936 + (-9.8t/2)

Vf = 5.936 + ((-9.8 * .251)/2)

Vf = 4.706 m/s

2)Magnitude = absolute value of Vf - Vo

Vo = 2(5.936) - 4.706

Vo = 7.166

|Vf-Vo|=Magnitude

|4.706 - 7.166| = 2.46 m/s

3) x = (Vf^2 - Vo^2)/2a

x = (0 - 4.706^2)/(2 * -9.8)

x = 1.12992

dx = Vo*t + .5 at^2

dx = 4.706t + .5 * -9.8t^2

1.12992 = 4.706t - 4.9t^2

1.12992 = t(4.706 - 4.9t)

1.12992/t = 4.706 - 4.9t

(1.12992/t) + 4.9t = 4.706

1.12992 + 4.9t = 4.706t

1.12992 = -.194t

t = -5.82455

Obviously time shouldn't be negative. Did I do parts 1 & 2 correctly? Also, where did I go wrong with part 3?

Thanks in advance for any help.