Question

10 numeric 1 point 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 range of the saxophone in meters. answer

Explanation:

Step1: Resolve initial velocity components

The initial horizontal velocity $v_{0x}=v_0\cos\theta$, and initial vertical velocity $v_{0y}=v_0\sin\theta$. Given $v_0 = 8\ m/s$ and $\theta = 40^{\circ}$, so $v_{0x}=8\cos40^{\circ}\approx 8\times0.766 = 6.128\ m/s$, $v_{0y}=8\sin40^{\circ}\approx8\times0.643 = 5.144\ m/s$.

Step2: Find the time - of - flight using vertical motion

The vertical displacement $y - y_0=- 15\ m$ (taking downwards as negative), and the equation for vertical displacement is $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, where $g = 9.8\ m/s^{2}$. So, $-15 = 5.144t-4.9t^{2}$. Rearranging gives $4.9t^{2}-5.144t - 15=0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $at^{2}+bt + c = 0$, here $a = 4.9$, $b=-5.144$, $c=-15$. Then $t=\frac{5.144\pm\sqrt{(-5.144)^{2}-4\times4.9\times(-15)}}{2\times4.9}=\frac{5.144\pm\sqrt{26.46+294}}{9.8}=\frac{5.144\pm\sqrt{320.46}}{9.8}=\frac{5.144\pm17.9}{9.8}$. We take the positive root $t=\frac{5.144 + 17.9}{9.8}=\frac{23.044}{9.8}\approx2.351\ s$.

Step3: Calculate the range

The range $R$ is given by the horizontal - motion equation $R = v_{0x}t$. Substituting $v_{0x}=6.128\ m/s$ and $t = 2.351\ s$, we get $R=6.128\times2.351\approx14.4\ m$.

Answer:

$14.4$