Physics
posted by Priscilla .
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
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