Question

9 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 time that the saxophone remains in the air before landing on the ground below. answer

Explanation:

Step1: Identify vertical motion variables

We analyze the vertical (y - direction) motion of the saxophone. The initial vertical velocity \(v_{0y}\) is given by \(v_{0y}=v_0\sin\theta\), where \(v_0 = 8\space m/s\) and \(\theta=30^{\circ}\). So \(v_{0y}=8\sin(30^{\circ})=8\times\frac{1}{2} = 4\space m/s\). The vertical displacement \(y=- 15\space m\) (negative because it is downward from the initial position), the acceleration \(a = - g=- 9.8\space m/s^{2}\) (negative because it is downward, we take upward as positive). We use the kinematic equation \(y=v_{0y}t+\frac{1}{2}at^{2}\).

Step2: Substitute values into the equation

Substitute \(y = - 15\), \(v_{0y}=4\) and \(a=-9.8\) into the equation \(y = v_{0y}t+\frac{1}{2}at^{2}\):
\(-15=4t+\frac{1}{2}(-9.8)t^{2}\)
Simplify the equation:
\(-15 = 4t-4.9t^{2}\)
Rearrange it to the standard quadratic form \(ax^{2}+bx + c = 0\):
\(4.9t^{2}-4t - 15=0\)

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 4.9\), \(b=-4\), \(c=-15\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-4)^{2}-4\times4.9\times(-15)=16 + 294=310\)
Then, \(t=\frac{4\pm\sqrt{310}}{2\times4.9}=\frac{4\pm17.61}{9.8}\)
We have two solutions:
\(t_1=\frac{4 + 17.61}{9.8}=\frac{21.61}{9.8}\approx2.205\space s\)
\(t_2=\frac{4-17.61}{9.8}=\frac{- 13.61}{9.8}\approx - 1.39\space s\)
Since time cannot be negative, we take the positive solution.

Answer:

\(\approx2.2\space s\) (or more precisely, using the quadratic formula calculation, we can get a more accurate value. If we calculate \(\frac{4+\sqrt{310}}{9.8}\), \(\sqrt{310}\approx17.6068\), then \(\frac{4 + 17.6068}{9.8}=\frac{21.6068}{9.8}\approx2.205\space s\), so the time is approximately \(2.2\space s\) (or \(2.21\space s\) depending on the level of precision))