A plane, diving at an angle of $37{}^\circ $ below the horizontal, releases a projectile at an altitude of $730\ m$. At the moment of release, the projectile and plane are traveling in the same direction at the same speed. The projectile hits the ground $5.0\ s$ after release.

1. What was the speed of the plane?

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We let $v_0$ denote the speed of the plane and $t_1$ denote the time at which the projectile hits the ground.

The kinematic equations of the projectile, after being released at $t=0$, are

\left\{ \begin{array}{c} x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \\ v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \end{array} \right.\ \ \ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2-v_0{\mathrm{sin} \left(\theta \right)\ }t+H} \\ v_y\left(t\right)=-gt-v_0{\mathrm{sin} \left(\theta \right)\ } \end{array} \right.

At time $t_1=5.0\ s$, the projectile reaches the ground and therefore $t_1$ satisfies $y\left(t_1\right)=0$. We then solve for $v_0$ as follows

\begin{aligned} y\left(t_1\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_1-v_0{\mathrm{sin} \left(\theta \right)\ }t_1+H=0 \\ &\Rightarrow \ \ \ \ \ v_0{\mathrm{sin} \left(\theta \right)\ }t_1=H-\frac{1}{2}gt^2_1 \\ &\Rightarrow \ \ \ \ \ v_0=\frac{\displaystyle{H-\frac{1}{2}gt^2_1}}{{\mathrm{sin} \left(\theta \right)\ }t_1} \end{aligned}

Thus, the speed of the plane is equal to

\boxed{v_0=\frac{\displaystyle{H-\frac{1}{2}gt^2_1}}{{\mathrm{sin} \left(\theta \right)\ }t_1}-\frac{\displaystyle{730-\frac{1}{2}\cdot 10\cdot 5^2}}{{\mathrm{sin} \left(37\right)\ }\cdot 5}\approx 202\ m/s}

2. How far does the projectile travel horizontally during its flight?

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The horizontal distance traveled by the projectile in the time $t_1$ is equal to

\boxed{R=x\left(t_1\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t_1=202\cdot {\mathrm{cos} \left(37\right)\ }\cdot 5\approx 807\ m}

3. What is the speed and direction of motion of the projectile just before hitting the ground? Express direction as an angle below the horizontal.

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The projectile hits the ground with a velocity ${\overrightarrow{v}}_1$ that has components $v_x\left(t_1\right)$ and $v_y\left(t_1\right)$ as shown below

The components are equal to

\begin{aligned} v_x\left(t_1\right)&=v_0{\mathrm{cos} \left(\theta \right)\ }=202\cdot {\mathrm{cos} \left(37\right)\ }\approx 161.3\ m/s \\ v_y\left(t_1\right)&=-gt_1+v_0{\mathrm{sin} \left(\theta \right)\ }=-9.8\cdot 5+202\cdot {\mathrm{sin} \left(37\right)\ }\approx -170.6\ \ m/s \end{aligned}

The speed of the projectile when it hits the ground is therefore equal to

\boxed{v_1=\sqrt{v_x{\left(t_1\right)}^2+v_y{\left(t_1\right)}^2}=\sqrt{{161.3}^2+{\left(-170.6\right)}^2}\approx 235\ \ m/s}

The direction of the velocity ${\overrightarrow{v}}_1$ is given by the angle $\beta $ which can be computed as follows

{\mathrm{tan} \left(\beta \right)\ }=\frac{v_y\left(t_1\right)}{v_x\left(t_1\right)}\ \ \ \ \ \Rightarrow \ \ \ \ \ \boxed{\beta ={{\mathrm{tan}}^{-1} \left(\frac{v_y\left(t_1\right)}{v_x\left(t_1\right)}\right)\ }={{\mathrm{tan}}^{-1} \left(-\frac{170.6}{161.3}\right)\ }\approx -46.6{}^\circ}

The projectile hits the ground at an angle $\beta \approx 46.6{}^\circ $ below the horizontal.

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