A basketball is shot from an initial height of 2.4 m with an initial speed (v0 = 12 m/s) directed at an angle (35 degrees) above the horizontal.
Additional information:
Height of the basket is 3.05 m
Find how far away the player is standing.
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To find how far away the player is standing, we need to use the principles of projectile motion. Here's what we can do step by step:
1. Break down the initial velocity into its vertical and horizontal components:
- The vertical component, vy, can be found by multiplying the initial speed (v0) by the sine of the launch angle (θ). So, vy = v0 * sin(θ).
- The horizontal component, vx, can be found by multiplying the initial speed (v0) by the cosine of the launch angle (θ). So, vx = v0 * cos(θ).
2. Calculate the time taken for the basketball to reach its peak height by using the equation:
- The time of flight, tf = vy / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
3. Calculate the time taken for the basketball to fall from its peak height to the height of the basket. Since the basketball started at a height of 2.4 m and the basket is at a height of 3.05 m, this height difference is given by:
- Δh = (height of the basket - initial height) = 0.65 m.
4. Calculate the time for the second part of the flight using the equation:
- Δt = sqrt(2 * Δh / g), where g is the acceleration due to gravity.
5. Find the total time of flight, t_total, by adding the times obtained in steps 2 and 4.
6. Calculate the horizontal distance covered, x, during the total flight time. Since the horizontal velocity remains constant throughout the flight, we can multiply the horizontal velocity component, vx, by the total time, t_total. So, x = vx * t_total.
7. The calculated distance from step 6 is the required answer.
Now, let's compute the values using the given information and equations:
- v0 = 12 m/s
- θ = 35 degrees
- Δh = 0.65 m
- g ≈ 9.8 m/s²
First, calculate the vertical and horizontal components using the formulas mentioned above:
vy = v0 * sin(θ) = 12 * sin(35°) ≈ 6.90 m/s
vx = v0 * cos(θ) = 12 * cos(35°) ≈ 9.82 m/s
Next, calculate the time to reach the peak height:
tf = vy / g = 6.90 / 9.8 ≈ 0.70 s
Then, calculate the time to fall from the peak height to the basket:
Δt = sqrt(2 * Δh / g) = sqrt(2 * 0.65 / 9.8) ≈ 0.34 s
Now, find the total time of flight:
t_total = tf + Δt = 0.70 + 0.34 ≈ 1.04 s
Finally, calculate the horizontal distance covered during the total flight time:
x = vx * t_total = 9.82 * 1.04 ≈ 10.19 m
Therefore, the player is approximately standing 10.19 meters away from the basket.