A basketball is launched with an initial speed of 12.0 m/s at 45 degrees from the ground.
The height of the basket y = 2.50m. From what distance x the ball is shot?
Note : The trajectory is like a basketball shot, the highest point the ball reaches is higher than the basket height.
vertical:
2.5=vo*t*sintheta-4.9t^2
2.5=12*t*.707 - 4.9t^2
Horizontal:
d=12*.707 t
Ok, solve for t in the second equation
t=d/(12*.707)
put it in the first equation
2.5=12(d/12*.707)*.707- 4.9 (2d^2/144)
2.5=d-9.8 d^2/144
solve that quadratic for d. which is x in your problem description
The value of d is coming out to be 3.194 and 11.499. Can both be the answers? If not, what value to eliminate and why?
To find the distance from which the basketball is shot (x), we need to analyze the vertical and horizontal components of motion separately.
Let's start by analyzing the vertical motion:
The basketball is shot at an angle of 45 degrees. This means that the initial velocity (v0) can be split into vertical and horizontal components.
Vertical component: v0y = v0 * sin(theta)
v0y = 12.0 m/s * sin(45°) (Using the trigonometric relationship: sin(45°) = √2 / 2)
v0y = 12.0 m/s * (√2 / 2)
v0y ≈ 8.49 m/s
Now, we can use this vertical component of velocity to find the time it takes for the basketball to reach its highest point above the ground (y = 2.50m). At the highest point, the velocity in the vertical direction becomes zero (vy = 0).
Using the kinematic equation for vertical motion:
v = v0 + at,
where v and v0 are the final and initial velocities, a is the acceleration, and t is the time.
At the highest point, the velocity in the vertical direction (vy) is zero. We can set v = 0 and solve for t.
0 = v0y + (-9.8 m/s^2) * t (taking the acceleration due to gravity as -9.8 m/s^2)
- v0y = -9.8 m/s^2 * t
t = v0y / 9.8 m/s^2
t ≈ (8.49 m/s) / (9.8 m/s^2)
t ≈ 0.867 s
Now that we have the time at which the basketball reaches its highest point (0.867 s), we can find the horizontal distance traveled (x).
Using the kinematic equation for horizontal motion:
x = v0x * t,
where v0x is the initial horizontal velocity, and t is the time.
Since the horizontal velocity remains constant throughout the motion, v0x can be calculated using:
v0x = v0 * cos(theta)
v0x = 12.0 m/s * cos(45°) (Using the trigonometric relationship: cos(45°) = √2 / 2)
v0x = 12.0 m/s * (√2 / 2)
v0x ≈ 8.49 m/s
Now, we can substitute the values into the equation to find the horizontal distance traveled (x).
x = (8.49 m/s) * (0.867 s)
x ≈ 7.36 m
Therefore, the basketball is shot from a distance of approximately 7.36 meters.
Note: The trajectory of the basketball is parabolic, so the highest point it reaches is higher than the basket height. Hence, we can still calculate the distance from which it is shot even though the highest point is higher than the basket.