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March 27, 2015

March 27, 2015

Posted by **Priscilla** on Wednesday, February 22, 2012 at 10:25am.

What horizontal distance does the ball cover before bouncing?

(delta)y = (Voy)(t) - (1/2)gt^2

(delta)y = (Vo)sin(teta)t - (1/2)gt^2

-0.80 = -4.9t^2 - 4.3sin15(t)

4.9t^2 + 4.3sin15(t) - 0.80 = 0

ax^2 + bx + c = 0

x = - (4.3sin15) +/- (sq root) (4.3sin15)^2 - 4 (4.9)(-.80)

divided by 2 (4.9)

x1 = .27

x2 = - .50

I know X2 cancels out because it's a negative number.

Then, I used

(delta)x = Vox x t

(delta)x = 4.3 m/s x .27 s

(delta)x = 1.161 m

But the answer is 1.3 m, I think I did something wrong in the quadratic equation. I'm really confused, thank you for any help

- Physics -
**drwls**, Wednesday, February 22, 2012 at 11:30am4.9 t^2 +1.113 t - 0.80 = 0

t = [-1.113 +sqrt(1.2386 +15.68]/9.8

= 3.002/9.8 = 0.3061 s

X = Vo *cos15*0.3061 = 1.27 m

You must have done the quadratic wrong, for t. You also forgot a cos15 factor for the horizontal motion.

- Physics -
**Priscilla**, Wednesday, February 22, 2012 at 3:52pmok thank you

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