Question

Question 615 marks You should be able to answer this question after studying Unit 10 At a battle re-enactment, a catapult launches a missile with a speed of 38 ms-, at an angle of 40° to the horizontal. The height from which the projectile is launched is 10m. Set up coordinate axes at the point where the missile leaves the catapult, as shown below 40° 10 m Let i and j be the Cartesian unit vectors in the positive directions of the r- and y-axes, respectively, as shown in the diagram Model the missile as a particle and assume that the only force acting on it is its weight. Take the magnitude of the acceleration due to gravity to be g 9.8ms 2. Give numerical answers to two significant figures (a) Show that the position r (in metres) of the missile t seconds after it has been aunched is given by r 38t cos 40i(38t sin40 2) j (b) Find the time (in seconds) that it takes for the missile to reach the ground (c) Determine the horizontal distance (in metres) between the point from where the missile is launched and the point where it first hits the ground

Answer 1

Set up the coordinate system so that the point from which the projectile is launched is the origin $0 \mathbf{i} + 0 \mathbf{j}$ .

(a) We use Newton's equations of motion to find the position of the missile.

$\mathbf{r} = \mathbf{r_0} + \mathbf{v}t + \frac{1}{2}\mathbf{a} t^2$ , where r₀ is the initial position, v is the velocity and a is the acceleration.

$\mathbf{r_0}$ is the initial position of the projectile, which is   $0 \mathbf{i} + 0 \mathbf{j}$ .

As the missile was launched at a speed of 38 ms^-1 at an angle of $40^{\circ }$ , we can find the horizontal and vertical components of the velocity $\mathbf{v}$ as $\mathbf{v} = 38 \cos (40) \mathbf{i} + 38 \sin (40) \mathbf{j}$ .

Now, the only force acting on the missile is gravity. So the acceleration is just the acceleration due to gravity in the negative vertical direction and 0 in the horizontal direction. In other words $\mathbf{a} = 0 \mathbf{i} - g \mathbf{j}$ .

If we substitute the above data in Newton's equation, we get

$\mathbf{r} = 0\mathbf{i} + 0\mathbf{j} + 38t \cos(40) \mathbf{i} + 38 t \sin(40) \mathbf{j} + 0 \mathbf{i} - \frac{1}{2}\mathbf{g} t^2 \mathbf{j}$ . Simplifying further, we find

$\mathbf{r} = 38t \cos(40) \mathbf{i} + 38 t \sin(40) \mathbf{j} - \frac{1}{2}\mathbf{g} t^2 \mathbf{j} \\ \mathbf{r} = 38t \cos(40) \mathbf{i} + \left (38 t \sin(40) - \frac{1}{2}\mathbf{g} t^2 \right ) \mathbf{j}$

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(b) Next, we want to find the time T it takes for the projectile to reach the ground.

Let R be the horizontal distance moved by the missile from launch to when it hits the ground. At the same time it also drops 10 m in height. So, when the missile hits the ground, its coordinates are

$\mathbf{r} = R \mathbf{i} - 10 \mathbf{j}$ . We substitute this into the formula found in (a).

$R \mathbf{i} - 10 \mathbf{j} = 38T \cos(40) \mathbf{i} + \left (38 T \sin(40) - \frac{1}{2}\mathbf{g} T^2 \right ) \mathbf{j}$

Equating the horizontal components gives us an equation involving horizontal range and T, which we cannot solve right now as it involves 2 unknown quantities.

Equating the vertical components gives us a quadratic equation in T, which we can solve.

$-10 = 38 T \sin(40) - \frac{1}{2} \times 9.8 \times T^2$ . Rearranging this equation, we find

$\frac{1}{2} \times 9.8 \times T^2 - \left ( 38 \sin (40) \right ) T - 10 = 0$

Applying the quadratic formula, we find 2 possible values of T. Of these, we take the positive value (the negative value corresponds to moving backwards in time).

$T = \frac{38 \sin (40) \pm \sqrt{\left ( 38 \sin (40) \right )^2 + 4 \times 10 \times \frac{9.8}{2}}}{9.8}$

$T = \frac{38 \sin (40) \pm \sqrt{\left ( 38 \sin (40) \right )^2 + 40 \times 4.9}}{9.8}$

Substituting sin (40) = 0.64278, we get

$T = \frac{ 24.4259 \pm \sqrt{\left( 24.4259 \right )^2 + 40 \times 4.9}}{9.8}$

$T = \frac{ 24.4259 \pm \sqrt{ 792.6260}}{9.8}$

$T = \frac{ 24.4259 \pm 28.15361}{9.8}$

Then, the required positive root is

$T = \frac{ 24.4259+28.15361}{9.8} = \frac{52.57954}{9.8} = 5.3652$ seconds.

Rounding off to 2 significant figures, we get T = 5.4 s.

The missile takes 5.4 s to reach the ground.

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(c) The next question asks us to find the horizontal distance traveled by the missile. In (b), we had equated vertical components to find T. If we now equate the horizontal components, we get

$R = 38 T \cos (40)$ . Since we found T above, we can just substitute the value of T to find R.

$R = 38 \times 5.4 \times \cos (40 ) = 38 \times 5.4 \times 0.76604 = 157.2$ meters.

The horizontal distance between the point of launch and the point where the missile first hits the ground is 157.2 m.