Math Problem Statement

She shoots the basketball with a speed of 58.1 ft/s while standing 48.5 ft away from the basketball hoop. The hoop is 10 ft above the floor and the basketball leaves her hand 6.3 ft above the floor. Determine the larger of the two angles that Samantha can shoot the basketball to win the prize. We know the smaller angle is 18.4 degrees

Solution

To determine the larger of the two angles at which Samantha can shoot the basketball, we can use the physics of projectile motion. Here's a step-by-step approach:

Given:

  • Initial speed (v0v_0): 58.1 ft/s
  • Horizontal distance to the hoop (dd): 48.5 ft
  • Height of the hoop (hh): 10 ft
  • Initial height of the ball (h0h_0): 6.3 ft
  • Smaller angle (θ1\theta_1): 18.4 degrees

Equations of Projectile Motion:

  1. Horizontal motion: d=v0cos(θ)td = v_0 \cdot \cos(\theta) \cdot t

  2. Vertical motion: h=h0+v0sin(θ)t12gt2h = h_0 + v_0 \cdot \sin(\theta) \cdot t - \frac{1}{2} g t^2

Where:

  • gg is the acceleration due to gravity (32.2ft/s232.2 \, \text{ft/s}^2).
  • θ\theta is the launch angle.
  • tt is the time of flight.

Step-by-Step Solution:

  1. Find time of flight (tt) in terms of θ\theta: Using the horizontal motion equation: t=dv0cos(θ)t = \frac{d}{v_0 \cos(\theta)}

  2. Substitute tt into the vertical motion equation: h=h0+v0sin(θ)(dv0cos(θ))12g(dv0cos(θ))2h = h_0 + v_0 \sin(\theta) \left(\frac{d}{v_0 \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{d}{v_0 \cos(\theta)}\right)^2

  3. Simplify the equation: 10=6.3+48.5tan(θ)16.1×(48.5)2(58.1)2cos2(θ)10 = 6.3 + 48.5 \tan(\theta) - \frac{16.1 \times (48.5)^2}{(58.1)^2 \cos^2(\theta)}

  4. Solve for θ\theta: This equation needs to be solved numerically for the unknown angle θ\theta. Given the smaller angle is 18.4 degrees, we solve for the larger angle using the trigonometric identity for projectiles. The two angles are complementary in terms of their tangent functions: θ2=90θ1\theta_2 = 90^\circ - \theta_1.

The equation involves a trigonometric function (tan(θ)\tan(\theta) and cos2(θ)\cos^2(\theta)), which is typically solved numerically. However, because we know the complementary nature of projectile angles:

θ2=90θ1\theta_2 = 90^\circ - \theta_1

Plugging the known smaller angle:

θ2=9018.4=71.6\theta_2 = 90^\circ - 18.4^\circ = 71.6^\circ

Thus, the larger angle at which Samantha can shoot the basketball is approximately 71.6 degrees.

Would you like a detailed breakdown of each step, or do you have any questions?

Related Questions:

  1. How would the angles change if Samantha were closer to or further from the hoop?
  2. What would be the effect of a different initial speed on the shooting angles?
  3. How does the difference in initial height of the ball affect the angles?
  4. How can you calculate the time of flight for each angle?
  5. What happens if the hoop height is increased or decreased?

Tip: Always ensure the units for speed, distance, and acceleration are consistent when applying physics equations.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Trigonometry
Kinematics

Formulas

Horizontal motion: d = v_0 * cos(θ) * t
Vertical motion: h = h_0 + v_0 * sin(θ) * t - (1/2) * g * t^2
Time of flight: t = d / (v_0 * cos(θ))
Trigonometric identity: θ_2 = 90° - θ_1

Theorems

Projectile Motion Theorem
Trigonometric Identities

Suitable Grade Level

Grades 10-12