Answer to Question #237055 in Physics for dnail

Question

A missile is detected flying with a speed of 580kmh-1 at an angle of 23o to the horizontal in mid-air 4.5 km above sea level.

Sketch a diagram indicating the position of the missile in mid-air. Hence or otherwise, solve the follows:

1. The launching speed

2. The launch angle.

3. The horizontal distance from the detected position to its launch.

4. The highest vertical displacement the missile can reach from sea level.

5. The time from the launch to its detection.

6. The horizontal distance from the launch to the landing (assumed not intercepted)

7. The total time in flight from launch to land

Answer 1

1. The launching speed

2. The launch angle

$$H=\frac{v_{yi}^2-v_{y_f}^2}{2g}=\frac{v_{yi}^2-(u\sin\theta)^2}{2g},\\\space\\ v_{yi}=\sqrt{2gH+(u\sin\theta)^2}=304\text{ m/s}.\\\space\\ v_{yi}=v\sin\alpha,\\ v_{xi}=v\cos\alpha=\text{const},\\ v_{xf}=u\cos\theta=\text{const},\\ v_{xf}=v_{xi},\\ u\cos\theta=v\cos\alpha.\\\space\\ \frac{v_{yi}}{u\cos\theta}=\frac{v\sin\alpha}{v\cos\alpha}=\tan\alpha,\\\space\\ \alpha=\arctan\frac{v_{yi}}{u\cos\theta}=\arctan\frac{304}{161\cos23°}=64°.\\\space\\ v=\frac{v_{yi}}{\sin\alpha}=\frac{304}{\sin64°}=338\text{ m/s}.$$

3. The horizontal distance from the detected position to its launch

$$t=\frac{v_{yi}-v_{yf}}{g}=\frac{v\sin\alpha-u\sin\theta}{g}=24.6\text{ s},\\\space\\ d=v_xt=vt\cos\alpha =3642\text{ m}.$$

4. The highest vertical displacement the missile can reach from sea level.

$$H_\text{max}=\frac{v_{yi}^2}{2g}=\frac{(v\sin\alpha)^2}{2g}=4709\text{ m}.$$

5. The time from the launch to its detection.

See point 3: $t=24.6\text{ s}.$

6. The horizontal distance from the launch to the landing (assumed not intercepted)

$$R=v_xT=v_x·2\frac{v_{yi}}{g}=\frac{v^2\sin(2\alpha)}{g}=9186\text{ m}.$$

7. The total time in flight from launch to land

$$T=2\frac{v_{yi}}{g}=2\frac{v\sin\alpha}{g}=62\text{ s}.$$