A basketball is launched with an initial speed of 10.0 m/s and follows the trajectory shown. The height of the basket y= 1.50m.From what distance x the ball is shot? Note: The drawing is not to scale. Angle is 45degrees
no diagram.
no info on initial height (hi) of ball
figure that out and we have
h(x) = hi + x - 9.8/100 x^2
solve for x when h = 1.5
To find the distance at which the basketball is shot, we can use the equations of projectile motion.
Given:
Initial speed (u) = 10.0 m/s
Height of the basket (y) = 1.50 m
Angle of launch (θ) = 45 degrees
First, we need to break the initial velocity into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
Horizontal component of velocity (ux) = u * cos(θ)
Vertical component of velocity (uy) = u * sin(θ)
By using the given angle (θ = 45 degrees) and the initial speed (u = 10.0 m/s), we can calculate the values for ux and uy as follows:
ux = 10.0 m/s * cos(45 degrees) ≈ 7.07 m/s
uy = 10.0 m/s * sin(45 degrees) ≈ 7.07 m/s
Next, we can calculate the time (t) it takes for the basketball to reach the height of the basket. At the maximum height, the vertical component of velocity is zero. We can use this information to find the time taken to reach that height.
The equation for finding the time of flight is:
t = uy / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
t = 7.07 m/s / 9.8 m/s² ≈ 0.72 s
Since the time of flight is the same for both the upward and downward motion, we can find the range (R) of the projectile:
R = ux * t
R = 7.07 m/s * 0.72 s ≈ 5.09 m
Therefore, the basketball is shot from a distance of approximately 5.09 meters.