Question

10 numeric 1 point a frustrated band student decides to rid herself of her saxophone once and for all to pursue a life dedicated to the french horn. she throws her saxophone from the top of a 15 meter tall cliff with an initial velocity of 8 m/s at an angle of 30 degrees from the horizontal. calculate the range of the saxophone in meters. answer

Explanation:

Step1: Break initial velocity into components

The initial velocity \( v_0 = 8 \, \text{m/s} \) at an angle \( \theta = 30^\circ \).
Horizontal component: \( v_{0x} = v_0 \cos\theta = 8 \cos30^\circ = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3} \, \text{m/s} \)
Vertical component: \( v_{0y} = v_0 \sin\theta = 8 \sin30^\circ = 8 \cdot \frac{1}{2} = 4 \, \text{m/s} \)

Step2: Analyze vertical motion (displacement \( y = -15 \, \text{m} \), acceleration \( a = -g = -9.8 \, \text{m/s}^2 \))

Use the equation \( y = v_{0y}t + \frac{1}{2}at^2 \).
Substitute values: \( -15 = 4t - 4.9t^2 \)
Rearrange: \( 4.9t^2 - 4t - 15 = 0 \)

Step3: Solve quadratic equation for time \( t \)

Quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4.9 \), \( b = -4 \), \( c = -15 \).
Discriminant: \( \Delta = (-4)^2 - 4(4.9)(-15) = 16 + 294 = 310 \)
\( t = \frac{4 \pm \sqrt{310}}{9.8} \). Take positive root: \( t \approx \frac{4 + 17.61}{9.8} \approx \frac{21.61}{9.8} \approx 2.205 \, \text{s} \)

Step4: Calculate horizontal range (constant velocity \( v_{0x} \))

Range \( R = v_{0x} \cdot t = 4\sqrt{3} \cdot 2.205 \approx 4 \cdot 1.732 \cdot 2.205 \approx 15.3 \, \text{m} \) (more precise: \( 4\sqrt{3} \approx 6.928 \), \( 6.928 \cdot 2.205 \approx 15.28 \approx 15.3 \))

Answer:

\( \approx 15.3 \) (or more precisely, solving the quadratic gives \( t \approx 2.204 \, \text{s} \), \( R = 4\sqrt{3} \cdot 2.204 \approx 15.27 \approx 15.3 \) meters)