A basketball player is standing on the floor 10.0 m from the basket as shown in the figure below. The height of the basket is H = 3.05 m, and he shoots the ball at an angle, 𜃠= 42.0°, with the horizontal from a height of h = 1.86 m. A) What is the magnitude of the acceleration, in m/s², of the basketball at the highest point in its trajectory? B) At what speed, in m/s, must the player throw the basketball so that the ball goes through the hoop without striking the backboard
The Correct Answer and Explanation is:
Part A: Magnitude of the acceleration at the highest point
At the highest point in the trajectory of a projectile, the vertical velocity component is zero. However, the horizontal component of velocity remains unchanged throughout the flight, because there are no horizontal forces acting on the projectile (assuming air resistance is negligible).
Since the only force acting on the basketball at all points during its flight is gravity, the acceleration at the highest point (and throughout the entire flight) is simply due to gravity. This acceleration is always:a=g=9.81 m/s2a = g = 9.81 \, \text{m/s}^2a=g=9.81m/s2
Thus, the magnitude of the acceleration of the basketball at the highest point in its trajectory is 9.81 m/s².
Part B: Speed to throw the ball so that it goes through the hoop
We need to use the kinematic equations to find the initial speed required for the ball to go through the hoop. The basketball player’s position, the height of the basket, and the angle of the shot give us the following information:
- Initial height of the shot: h=1.86 mh = 1.86 \, \text{m}h=1.86m
- Height of the hoop: H=3.05 mH = 3.05 \, \text{m}H=3.05m
- Distance to the hoop: d=10.0 md = 10.0 \, \text{m}d=10.0m
- Angle of launch: θ=42.0∘\theta = 42.0^\circθ=42.0∘
To find the required initial speed v0v_0v0, we need to break the motion into horizontal and vertical components. The equations for the motion are:
1. Horizontal motion:x(t)=v0cos(θ)⋅tx(t) = v_0 \cos(\theta) \cdot tx(t)=v0cos(θ)⋅t
At the time the ball reaches the basket (horizontal distance x=10.0 mx = 10.0 \, \text{m}x=10.0m):10.0=v0cos(42∘)⋅t10.0 = v_0 \cos(42^\circ) \cdot t10.0=v0cos(42∘)⋅t
Solve for time ttt:t=10.0v0cos(42∘)t = \frac{10.0}{v_0 \cos(42^\circ)}t=v0cos(42∘)10.0
2. Vertical motion:y(t)=h+v0sin(θ)⋅t−12gt2y(t) = h + v_0 \sin(\theta) \cdot t – \frac{1}{2} g t^2y(t)=h+v0sin(θ)⋅t−21gt2
At the time the ball reaches the basket (vertical distance y=3.05 my = 3.05 \, \text{m}y=3.05m):3.05=1.86+v0sin(42∘)⋅t−12gt23.05 = 1.86 + v_0 \sin(42^\circ) \cdot t – \frac{1}{2} g t^23.05=1.86+v0sin(42∘)⋅t−21gt2
Substitute t=10.0v0cos(42∘)t = \frac{10.0}{v_0 \cos(42^\circ)}t=v0cos(42∘)10.0 into this equation:3.05=1.86+v0sin(42∘)⋅10.0v0cos(42∘)−12g(10.0v0cos(42∘))23.05 = 1.86 + v_0 \sin(42^\circ) \cdot \frac{10.0}{v_0 \cos(42^\circ)} – \frac{1}{2} g \left( \frac{10.0}{v_0 \cos(42^\circ)} \right)^23.05=1.86+v0sin(42∘)⋅v0cos(42∘)10.0−21g(v0cos(42∘)10.0)2
Simplify this equation:3.05=1.86+10.0tan(42∘)−12g(10.02v02cos2(42∘))3.05 = 1.86 + 10.0 \tan(42^\circ) – \frac{1}{2} g \left( \frac{10.0^2}{v_0^2 \cos^2(42^\circ)} \right)3.05=1.86+10.0tan(42∘)−21g(v02cos2(42∘)10.02)
Now solve for v0v_0v0. This requires algebraic manipulation, but using the equation solver or a numerical approach, we find that:v0≈8.52 m/sv_0 \approx 8.52 \, \text{m/s}v0≈8.52m/s
Final Answer:
- A) The magnitude of the acceleration at the highest point is 9.81 m/s29.81 \, \text{m/s}^29.81m/s2.
- B) The speed at which the player must throw the ball is approximately 8.52 m/s8.52 \, \text{m/s}8.52m/s.
