A basketball hoop is 10 feet high; and the distance from the basket to the free throw line is 15 feet: While shooting a free throw, a player releases the ball 7 feet from the ground at an angle of 0 40 degrees with respect to horizontal. What is the magnitude of the ball’s velocity needed to make a basket? You can neglect the width of the basketball Use 1 foot 0.3 meters. Make sure to keep plenty of significant figures in the intermediate stages of your calculation: 15 feet 10 feet 7 feet
The Correct Answer and Explanation is:
Solution Overview:
We are asked to find the magnitude of the ball’s initial velocity needed to make a basketball shot using projectile motion principles. The player releases the ball at a height of 7 feet, and it must reach a basket at 10 feet high, 15 feet away horizontally, at an angle of θ=40∘\theta = 40^\circ.
Given (Converted to SI Units):
- Initial height: y0=7 ft=2.1 my_0 = 7 \text{ ft} = 2.1 \text{ m}
- Final height: y=10 ft=3.0 my = 10 \text{ ft} = 3.0 \text{ m}
- Horizontal distance: x=15 ft=4.5 mx = 15 \text{ ft} = 4.5 \text{ m}
- Launch angle: θ=40∘\theta = 40^\circ
- Acceleration due to gravity: g=9.8 m/s2g = 9.8 \text{ m/s}^2
Equations of Motion:
From projectile motion:
- x=v0cos(θ)⋅tx = v_0 \cos(\theta) \cdot t → Solve for time tt: t=xv0cos(θ)t = \frac{x}{v_0 \cos(\theta)}
- Vertical displacement: y=y0+v0sin(θ)⋅t−12gt2y = y_0 + v_0 \sin(\theta) \cdot t – \frac{1}{2} g t^2
Substitute tt from (1) into (2): y=y0+v0sin(θ)⋅(xv0cos(θ))−12g(xv0cos(θ))2y = y_0 + v_0 \sin(\theta) \cdot \left( \frac{x}{v_0 \cos(\theta)} \right) – \frac{1}{2} g \left( \frac{x}{v_0 \cos(\theta)} \right)^2
Simplify: y=y0+xtan(θ)−gx22v02cos2(θ)y = y_0 + x \tan(\theta) – \frac{g x^2}{2 v_0^2 \cos^2(\theta)}
Now plug in values: 3.0=2.1+4.5tan(40∘)−9.8⋅(4.5)22v02cos2(40∘)3.0 = 2.1 + 4.5 \tan(40^\circ) – \frac{9.8 \cdot (4.5)^2}{2 v_0^2 \cos^2(40^\circ)}
- tan(40∘)≈0.8391\tan(40^\circ) ≈ 0.8391
- cos(40∘)≈0.7660\cos(40^\circ) ≈ 0.7660
- cos2(40∘)≈0.5868\cos^2(40^\circ) ≈ 0.5868
3.0=2.1+3.776−9.8⋅20.252v02⋅0.58683.0 = 2.1 + 3.776 – \frac{9.8 \cdot 20.25}{2 v_0^2 \cdot 0.5868} 3.0=5.876−198.451.1736v023.0 = 5.876 – \frac{198.45}{1.1736 v_0^2} −2.876=−198.451.1736v02⇒v02=198.451.1736⋅2.876≈58.68⇒v0≈58.68≈7.66 m/s-2.876 = – \frac{198.45}{1.1736 v_0^2} \Rightarrow v_0^2 = \frac{198.45}{1.1736 \cdot 2.876} ≈ 58.68 \Rightarrow v_0 ≈ \sqrt{58.68} ≈ 7.66 \text{ m/s}
✅ Final Answer:
v0≈7.66 m/s\boxed{v_0 \approx 7.66 \ \text{m/s}}
Explanation
To calculate the velocity required for a basketball to reach the hoop, we apply principles from two-dimensional projectile motion. The ball is launched at a 40° angle above the horizontal, from a height of 7 feet (2.1 m), and must reach a hoop that is 10 feet (3.0 m) high and 15 feet (4.5 m) away.
Using the horizontal motion equation, we express the time of flight in terms of the unknown initial velocity and known distance. This time expression is substituted into the vertical motion equation, which incorporates the gravitational pull acting downward. The vertical equation models how the ball’s height changes with time, accounting for the initial height, vertical velocity component, and gravity.
By combining these equations, we eliminate time and create a relationship solely involving the known quantities and the unknown initial velocity. After simplifying and substituting trigonometric values for 40°, we solve algebraically for the square of the initial velocity and then take the square root to find its magnitude.
This approach assumes ideal conditions: no air resistance, a point-like ball, and perfect aiming. Despite these simplifications, the result provides a reliable estimate. The final required velocity is approximately 7.66 m/s, a realistic value for a skilled basketball player taking a free throw.
