# Physics

posted by on .

In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.3 m/s at an angle of 15 degrees below the horizontal. It is released 0.80m above the floor.

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 - ,

4.9 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 - ,

ok thank you