In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 3.9 m/s at an angle of 13.8° below the horizontal. It is released 0.65 m above the floor. What horizontal distance does the ball cover before bouncing?
it's initial vertical downward speed is 3.9sin13.8
h=.65 -3.9sin13.8*t-4.9t^2=0 solve for t
then, horizontal distance=3.9cos13.8*t
How high is the ball when it hits the wall?
How long does it take for the ball to reach the wall if it is 5.2 m away?
To find the horizontal distance the ball covers before bouncing, we can use the kinematic equations of motion.
First, let's determine the time it takes for the ball to hit the floor after being released. We can use the vertical motion equation:
y = y0 + v0y*t - (1/2)*g*t^2
Where:
y = final vertical position (0 m, since the ball hits the floor)
y0 = initial vertical position (0.65 m)
v0y = initial vertical velocity (v0 * sin(θ))
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken
Since the ball is thrown vertically downward, the initial vertical velocity (v0y) would be negative. Thus,
0 = 0.65 m + (-(3.9 m/s) * sin(13.8°)) * t - (1/2) * (9.8 m/s^2) * t^2
We can rearrange this equation to solve for t:
(1/2) * (9.8 m/s^2) * t^2 - (3.9 m/s) * sin(13.8°) * t - 0.65 m = 0
This is a quadratic equation in the form of at^2 + bt + c = 0, where:
a = (1/2) * (9.8 m/s^2)
b = -(3.9 m/s) * sin(13.8°)
c = -0.65 m
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Plugging in the values, we get:
t = (-(3.9 m/s) * sin(13.8°) ± √((-(3.9 m/s) * sin(13.8°))^2 - 4 * (1/2) * (9.8 m/s^2) * (-0.65 m)))/(2 * (1/2) * (9.8 m/s^2))
Calculating the values within the square root:
√((-(3.9 m/s) * sin(13.8°))^2 - 4 * (1/2) * (9.8 m/s^2) * (-0.65 m))
=√((-(3.9 m/s))^2 * (sin(13.8°))^2 + (9.8 m/s^2) * 0.65 m)
=√((15.21 m^2/s^2) * (0.0300) + (6.37 m^2/s^2))
Calculating the values within the parentheses:
√(15.21 m^2/s^2 * 0.0300 + 6.37 m^2/s^2)
= √(0.4563 m^2/s^2 + 6.37 m^2/s^2)
= √(6.8263 m^2/s^2)
Simplifying the square root:
√(6.8263 m^2/s^2) ≈ 2.61 m/s
We can substitute these values back into the formula for t:
t = (-(3.9 m/s) * sin(13.8°) ± 2.61 m/s) / (2 * (1/2) * (9.8 m/s^2))
There will be two possible values for t, one positive and one negative. Since time cannot be negative in this case, we discard the negative value. Therefore,
t = (-(3.9 m/s) * sin(13.8°) + 2.61 m/s) / (2 * (1/2) * (9.8 m/s^2))
Now that we have calculated the time it takes for the ball to hit the floor, we can determine the horizontal distance covered by the ball. We can use the horizontal motion equation:
x = x0 + v0x * t
Where:
x = horizontal distance covered
x0 = initial horizontal position (0 m)
v0x = initial horizontal velocity (v0 * cos(θ))
t = time taken
In this case, the initial horizontal velocity (v0x) is given by
v0x = v0 * cos(θ)
= (3.9 m/s) * cos(13.8°)
Now we can substitute the values into the equation:
x = (0 m) + ((3.9 m/s) * cos(13.8°)) * ((-(3.9 m/s) * sin(13.8°) + 2.61 m/s) / (2 * (1/2) * (9.8 m/s^2)))
Simplifying further:
x = ((3.9 m/s) * cos(13.8°) * (-(3.9 m/s) * sin(13.8°) + 2.61 m/s)) / (2 * (1/2) * (9.8 m/s^2))
Evaluating this expression will give us the horizontal distance covered by the ball before bouncing.