A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as shown. If they shoot the ball at a 40.0° angle with the horizontal, and the basket height is 3.05 m, a. at what initial speed must they shoot so that it goes through the hoop without striking the backboard? b. find the velocity of the ball as it goes through the hoop. State the result as a vector.
The Correct Answer and Explanation is:
To solve this problem, we need to break it down into parts: the horizontal and vertical motions of the basketball, since projectile motion consists of these two independent motions. Let’s assume there is no air resistance.
Part a: Finding the initial speed
We can use the projectile motion equations for horizontal and vertical motion. We know the following:
- y0=2.00 my_0 = 2.00 \, \text{m}y0=2.00m (initial height of the basketball player)
- yf=3.05 my_f = 3.05 \, \text{m}yf=3.05m (height of the basket)
- x=10.0 mx = 10.0 \, \text{m}x=10.0m (horizontal distance from the player to the basket)
- θ=40.0∘\theta = 40.0^\circθ=40.0∘ (launch angle)
- g=9.81 m/s2g = 9.81 \, \text{m/s}^2g=9.81m/s2 (acceleration due to gravity)
Step 1: Horizontal motion equation
The horizontal distance traveled by the ball is given by:x=v0⋅cos(θ)⋅tx = v_0 \cdot \cos(\theta) \cdot tx=v0⋅cos(θ)⋅t
Where:
- v0v_0v0 is the initial velocity
- ttt is the time taken to reach the basket
- cos(θ)\cos(\theta)cos(θ) is the cosine of the launch angle
From this equation, we can solve for time ttt:t=xv0⋅cos(θ)t = \frac{x}{v_0 \cdot \cos(\theta)}t=v0⋅cos(θ)x
Step 2: Vertical motion equation
The vertical displacement of the ball is described by:yf=y0+v0⋅sin(θ)⋅t−12gt2y_f = y_0 + v_0 \cdot \sin(\theta) \cdot t – \frac{1}{2} g t^2yf=y0+v0⋅sin(θ)⋅t−21gt2
Substitute the expression for ttt from the horizontal motion equation into this vertical motion equation:yf=y0+v0⋅sin(θ)⋅xv0⋅cos(θ)−12g(xv0⋅cos(θ))2y_f = y_0 + v_0 \cdot \sin(\theta) \cdot \frac{x}{v_0 \cdot \cos(\theta)} – \frac{1}{2} g \left( \frac{x}{v_0 \cdot \cos(\theta)} \right)^2yf=y0+v0⋅sin(θ)⋅v0⋅cos(θ)x−21g(v0⋅cos(θ)x)2
Simplifying:yf=y0+x⋅tan(θ)−g⋅x22⋅v02⋅cos2(θ)y_f = y_0 + x \cdot \tan(\theta) – \frac{g \cdot x^2}{2 \cdot v_0^2 \cdot \cos^2(\theta)}yf=y0+x⋅tan(θ)−2⋅v02⋅cos2(θ)g⋅x2
Now, plug in the known values and solve for v0v_0v0:3.05=2.00+10.0⋅tan(40∘)−9.81⋅(10.0)22⋅v02⋅cos2(40∘)3.05 = 2.00 + 10.0 \cdot \tan(40^\circ) – \frac{9.81 \cdot (10.0)^2}{2 \cdot v_0^2 \cdot \cos^2(40^\circ)}3.05=2.00+10.0⋅tan(40∘)−2⋅v02⋅cos2(40∘)9.81⋅(10.0)2
After solving for v0v_0v0, we get:v0≈12.15 m/sv_0 \approx 12.15 \, \text{m/s}v0≈12.15m/s
So, the initial speed required to make the shot is approximately 12.15 m/s.
Part b: Finding the velocity of the ball at the hoop
To find the velocity of the ball as it reaches the basket, we need to determine both the horizontal and vertical components of velocity at that point.
Step 1: Horizontal velocity component
The horizontal velocity remains constant throughout the flight (since there is no air resistance):vx=v0⋅cos(θ)v_{x} = v_0 \cdot \cos(\theta)vx=v0⋅cos(θ)vx=12.15⋅cos(40∘)≈9.29 m/sv_{x} = 12.15 \cdot \cos(40^\circ) \approx 9.29 \, \text{m/s}vx=12.15⋅cos(40∘)≈9.29m/s
Step 2: Vertical velocity component
The vertical velocity changes due to gravity. The vertical velocity at the time of reaching the basket can be calculated using:vy=v0⋅sin(θ)−g⋅tv_{y} = v_0 \cdot \sin(\theta) – g \cdot tvy=v0⋅sin(θ)−g⋅t
Substituting the expression for ttt and solving for vyv_{y}vy, we get:vy=12.15⋅sin(40∘)−9.81⋅10.012.15⋅cos(40∘)v_{y} = 12.15 \cdot \sin(40^\circ) – 9.81 \cdot \frac{10.0}{12.15 \cdot \cos(40^\circ)}vy=12.15⋅sin(40∘)−9.81⋅12.15⋅cos(40∘)10.0vy≈7.81−4.79≈3.02 m/sv_{y} \approx 7.81 – 4.79 \approx 3.02 \, \text{m/s}vy≈7.81−4.79≈3.02m/s
Step 3: Total velocity
The total velocity is the vector sum of the horizontal and vertical components:v=vx2+vy2v = \sqrt{v_x^2 + v_y^2}v=vx2+vy2v=(9.29)2+(3.02)2≈9.67 m/sv = \sqrt{(9.29)^2 + (3.02)^2} \approx 9.67 \, \text{m/s}v=(9.29)2+(3.02)2≈9.67m/s
So, the total velocity of the ball as it goes through the hoop is approximately 9.67 m/s.
Conclusion:
- The initial speed required for the basketball player to shoot the ball is 12.15 m/s.
- The velocity of the ball as it goes through the hoop is 9.67 m/s at an angle of approximately 18.4° above the horizontal.
