practice it use the worked example above to help you solve this problem. an alaskan rescue plane drops a…

practice it use the worked example above to help you solve this problem. an alaskan rescue plane drops a package of emergency rations to stranded hikers, as shown in the figure. the plane is traveling horizontally at 35.0 m/s at a height of 1.50×10² m above the ground. (a) where does the package strike the ground relative to the point at which it was released? your response differs from the correct answer by more than 10%. double check your calculations. m (b) what are the horizontal and vertical components of the velocity of the package just before it hits the ground? horizontal m/s vertical your response differs from the correct answer by more than 10%. double check your calculations. m/s (c) find the angle of the impact. your response differs from the correct answer by more than 10%. double check your calculations. °

practice it use the worked example above to help you solve this problem. an alaskan rescue plane drops a package of emergency rations to stranded hikers, as shown in the figure. the plane is traveling horizontally at 35.0 m/s at a height of 1.50×10² m above the ground. (a) where does the package strike the ground relative to the point at which it was released? your response differs from the correct answer by more than 10%. double check your calculations. m (b) what are the horizontal and vertical components of the velocity of the package just before it hits the ground? horizontal m/s vertical your response differs from the correct answer by more than 10%. double check your calculations. m/s (c) find the angle of the impact. your response differs from the correct answer by more than 10%. double check your calculations. °

Answer

Explanation:

Step1: Calculate time of fall

The vertical - motion of the package is a free - fall. Using the equation $y = y_0+v_{0y}t-\frac{1}{2}gt^{2}$, where $y = 0$, $y_0=1.50\times 10^{2}\ m$, $v_{0y}=0\ m/s$. So, $0 = 1.50\times 10^{2}+0\times t - \frac{1}{2}\times9.8\ t^{2}$. Solving for $t$: [t=\sqrt{\frac{2y_0}{g}}=\sqrt{\frac{2\times1.50\times 10^{2}}{9.8}}] [t=\sqrt{\frac{300}{9.8}}\approx5.53\ s]

Step2: Calculate horizontal distance

The horizontal motion is a uniform - motion with $v_x = 35.0\ m/s$ and $t$ from Step 1. Using the equation $x = v_x t$, we have $x=35.0\times5.53 = 193.55\ m$

Step3: Calculate vertical component of velocity just before hitting the ground

Using the equation $v_y=v_{0y}-gt$, with $v_{0y} = 0\ m/s$, $g = 9.8\ m/s^{2}$ and $t = 5.53\ s$. So, $v_y=0 - 9.8\times5.53=- 54.194\ m/s$

Step4: Calculate angle of impact

The angle of impact $\theta$ is given by $\tan\theta=\frac{v_y}{v_x}$. Substituting $v_x = 35.0\ m/s$ and $v_y=-54.194\ m/s$, we get $\theta=\arctan\left(\frac{-54.194}{35.0}\right)\approx - 57.1^{\circ}$

Answer:

(a) $193.55\ m$ (b) horizontal: $35.0\ m/s$, vertical: $- 54.194\ m/s$ (c) $-57.1^{\circ}$