Question

a frustrated band student decides to rid herself saxophone once and for all to pursue a life dedic the french horn. she throws her saxophone from the top of a 15 m tall cliff with an initial velocity of 8 m/s at an ang degrees from the horizontal. calculate the time that the saxophone remains i air before landing on the ground below.

Explanation:

Step1: Resolve initial velocity into vertical - component

The initial vertical velocity $v_{0y}=v_0\sin\theta$, where $v_0 = 8\ m/s$ and assume $\theta = 30^{\circ}$, so $v_{0y}=8\times\sin30^{\circ}=8\times\frac{1}{2}=4\ m/s$. The height of the cliff $h = 15\ m$, and the vertical - displacement formula is $y = y_0+v_{0y}t-\frac{1}{2}gt^2$. Here $y = 0$, $y_0=15\ m$, $g = 9.8\ m/s^2$. So the equation becomes $0 = 15+4t-4.9t^2$.

Step2: Solve the quadratic equation

The quadratic formula for $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For the equation $- 4.9t^{2}+4t + 15=0$, where $a=-4.9$, $b = 4$, and $c = 15$. First, calculate the discriminant $\Delta=b^{2}-4ac=(4)^{2}-4\times(-4.9)\times15=16 + 294=310$. Then $t=\frac{-4\pm\sqrt{310}}{-9.8}$.
We take the positive root since time cannot be negative. $t=\frac{-4+\sqrt{310}}{-9.8}\approx\frac{-4 + 17.6}{-9.8}$ (we ignore the negative - root $t=\frac{-4-\sqrt{310}}{-9.8}$ as it gives a non - physical negative time value). $t=\frac{-4 + 17.6}{-9.8}=\frac{13.6}{9.8}\approx1.39\ s$.

Answer:

$1.39\ s$