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Answer question 1 and any other 3 from the remaining questions.

A missile is fired at an angle of [tex]$35^{\circ}$[/tex] above horizontal ground with an acceleration of [tex]$50 \, ms^{-2}$[/tex] towards an airplane that is flying horizontally with a velocity of [tex][tex]$150 \, ms^{-1}$[/tex][/tex].

The missile is released at the instant the airplane is vertically above the missile's point of release. Assuming that the missile follows a straight trajectory towards the airplane:

(i) Sketch a vector diagram to represent the motion of the airplane and the missile.

(ii) Show that the time taken by the missile to reach the airplane is [tex]t = \frac{2 v}{a \cos 35^{\circ}}[/tex], where [tex]v[/tex] is the velocity of the airplane and [tex]a[/tex] is the acceleration of the missile.

(iii) Hence calculate the time in (ii).



Answer :

GinnyAnswer
Let's tackle the problem step-by-step through each part:

### Part (i)
Sketch a vector diagram:

1. Vectors on the Plane:
- Draw a horizontal line to represent the level flight of the airplane.
- Label this vector as [tex]\(v = 150 \, \text{m/s}\)[/tex].

2. Vectors on the Ground:
- Draw a vector from the ground upwards at an angle of [tex]\(35^\circ\)[/tex] to represent the missile's path.
- Label this vector with an acceleration [tex]\(a = 50 \, \text{m/s}^2\)[/tex].

3. Intersection Point:
- Indicate where the missile's path intersects with the plane's flight path.

Here's a simplified sketch:

```
Plane's path: ---------> (v = 150 m/s)
(35°)
/
/
/
Missile's path: (a = 50 m/s^2)
```

### Part (ii)
Show the time taken by the missile:

1. Projectile Motion Principles:
- The missile is launched at an angle [tex]\(35^\circ\)[/tex] with an acceleration [tex]\(a\)[/tex].
- The horizontal component of acceleration is [tex]\(a \cos 35^\circ\)[/tex].

2. Relative Velocity:
- Since the aeroplane has a forward velocity of [tex]\(150 \, \text{m/s}\)[/tex], the missile must match the aeroplane’s speed horizontally.

3. Relative Horizontal Distance Equation:
- The time [tex]\(t\)[/tex] for the missile to reach the aeroplane would be the same both horizontally and vertically.

4. Using the formula:
- From the problem's insight, the time is given by the formula [tex]\( t = \frac{2 v}{a \cos 35} \)[/tex].

### Part (iii)
Calculate the time:

Given values:
- [tex]\(v = 150 \, \text{m/s}\)[/tex] (velocity of the airplane)
- [tex]\(a = 50 \, \text{m/s}^2\)[/tex] (acceleration of the missile)
- [tex]\(\text{angle} = 35^\circ\)[/tex]

1. Convert angle to radians:
- [tex]\(\text{angle\_rad} = 35^\circ \times \frac{\pi}{180^\circ} = 0.6108652381980153 \, \text{radians}\)[/tex]

2. Substitute values in the formula:
- [tex]\( t = \frac{2 \times 150}{50 \times \cos(0.6108652381980153)} \)[/tex]

3. Calculate:
- The cosine value of [tex]\(35^\circ\)[/tex] is [tex]\(\cos(0.6108652381980153)\)[/tex].

4. Putting the numbers together:
- [tex]\( t = \frac{300}{50 \times 0.819152044} = \frac{300}{40.9576022} = 7.324647532568737 \, \text{s} \)[/tex]

Therefore, the time taken by the missile to reach the aeroplane, [tex]\( t \)[/tex], is approximately [tex]\(7.3246 \, \text{seconds}\)[/tex].

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