example problem: an alaskan rescue plane drops a package of emergency rations to a stranded party of…

Explanation:

Step1: Calculate time of fall

The vertical - motion is a free - fall. Using the equation $y = y_0+v_{0y}t-\frac{1}{2}gt^{2}$, where $y = 0$, $y_0 = 100$ m, $v_{0y}=0$ m/s. So, $0 = 100- \frac{1}{2}(9.8)t^{2}$. Solving for $t$: $$t=\sqrt{\frac{2y_0}{g}}=\sqrt{\frac{2\times100}{9.8}}\ s\approx4.52\ s$$

Step2: Calculate horizontal distance

The horizontal motion is a uniform - motion with $v_x = 40.0$ m/s. Using the equation $x = v_x t$, substituting $v_x = 40.0$ m/s and $t = 4.52$ s, we get $x=40\times4.52 = 180.8$ m.

Step3: Find horizontal and vertical components of velocity

The horizontal component of velocity remains constant during the motion, so $v_x=40.0$ m/s. The vertical component of velocity is given by $v_y=v_{0y}-gt$. Since $v_{0y} = 0$ m/s and $t = 4.52$ s, $g = 9.8$ m/s², then $v_y=-9.8\times4.52=-44.3$ m/s.

Step4: Calculate the angle of impact

The angle of impact $\theta$ is given by $\tan\theta=\frac{v_y}{v_x}$. Substituting $v_x = 40.0$ m/s and $v_y=-44.3$ m/s, we get $\theta=\arctan\left(\frac{-44.3}{40.0}\right)\approx - 47.9^{\circ}$

Answer:

(a) $180.8$ m (b) $v_x = 40.0$ m/s, $v_y=-44.3$ m/s (c) $\theta\approx - 47.9^{\circ}$