10DProjectileMotion


 * Projectile Motion**

In this section, we will assume no air resistance. Therefore the only force acting on a projectile in flight will be gravity.

We take the acceleration due to gravity to be g = 9.8 m/s 2.

We consider motion in a vertical plane so we will define **__//i//__** to be horizontal and __**j**__ to be vertical.

... ... __**a**__(t) = -g __j__

**Example 1**

An object is pushed from the top of a cliff which is 500m high. Its initial horizontal speed is 10m/s. How far from the base of the cliff will the object land?

math \\ . \qquad \underline{r}(0)=0\underline{i}+500\underline{j} \\. \\ . \qquad \underline{v}(0)=10\underline{i}+0\underline{j} math
 * From the question **

math \\ . \qquad \underline{a}(t)=-g\underline{j} \\. \\ . \qquad \underline{v}(t)=\int -g\underline{j} \; dt \\. \\ . \qquad \underline{v}(t)=-gt\underline{j}+\underline{c} math
 * Start with acceleration = g downwards **

math \\ .\qquad \underline{v}(0)=10\underline{i}+0\underline{j} \quad \text{so} \\. \\ . \qquad \underline{c}=10\underline{i}+0\underline{j} \quad \text{so} \\ .\\ . \qquad \underline{v}(t)=10\underline{i}-gt\underline{j} math
 * but we have initial conditions for __v__(0) **

math \\ . \qquad \underline{r}(t)=\int 10\underline{i}-gt\underline{j} \; dt \\. \\ . \qquad \underline{r}(t)=10t\underline{i}-\frac{1}{2}gt^2\underline{j}+\underline{d} math
 * so now find position **

math \\ . \qquad \underline{r}(0)=0\underline{i}+500\underline{j} \quad \text{so}\\. \\ . \qquad \underline{d}=500\underline{j} \quad \text{so}\\. \\ . \qquad \underline{r}(t)=10t\underline{i}+\big( 500-\frac{1}{2}gt^2 \big) \underline{j} math
 * but we have initial conditions for __r__(0) **

math \\ . \qquad 500-\frac{1}{2}gt^2=0 \\. \\ . \qquad gt^2=1000 \\. \\ . \qquad t=\sqrt{\frac{1000}{g} } math
 * The object hits the ground when j (vertical) component of the position is zero. **

math \\ . \qquad t \approx 10.1 \; \text{seconds} \\. \\ . \qquad \underline{r}(10.1)=101\underline{i}+0\underline{j} math
 * ignore negative because t >= 0 **

So the object hits the ground, 101m from the base of the cliff.

**Initial Speed and Angle** In many problems, the initial velocity is provided in the form of speed, **v**, and angle of projection, **q **, where theta is the angle above the horizontal.

math . \qquad \underline{v}(t=0)=v\cos\theta\underline{i}+v\sin\theta\underline{j} math
 * The initial vector is therefore: **

**Example 2** A stone is projected with a speed of 50m/s at an angle of theta to the horizontal, where: math . \qquad \tan\theta=\frac{3}{4} math


 * Solution: **

math \\ . \qquad \sin\theta=\frac{3}{5} \\. \\ . \qquad \cos\theta=\frac{4}{5} math
 * From basic trigonometry we get that: **

math \\ . \qquad \underline{v}(0)=50\cos\theta\underline{i}+50\sin\theta\underline{j} \\ .\\ . \qquad \underline{v}(0)=40\underline{i}+30\underline{j} math
 * Thus, given v(0)=50 **


 * a) Find the vector for the velocity at time, t **

math \\ . \qquad \underline{a}(t)=-g\underline{j} \\ .\\ . \qquad \underline{v}(t)=\int-g\underline{j} \; dt \\ .\\ . \qquad \underline{v}(t)=-gt\underline{j} + \underline{c} math

math \\ . \qquad \underline{v}(0)=40\underline{i}+30\underline{j} \quad \text{so} \\ .\\ . \qquad \underline{c}=40\underline{i}+30\underline{j} \quad \text{so} \\ .\\ . \qquad \underline{v}(t)=40\underline{i}+\big( 30-gt \big) \underline{j} math
 * but we have initial conditions v(0) **


 * b) Find the vector for the position at time, t **

math \\ . \qquad \underline{r}(t)=\int \, 40\underline{i}+\big( 30-gt \big) \underline{j} \; dt \\ .\\ . \qquad \underline{r}(t)=40t\underline{i}+\big( 30t-\frac{1}{2}gt^2 \big) \underline{j} +\underline{d} math

math . \qquad \underline{r}(t)=40t\underline{i}+\big( 30t-\frac{1}{2}gt^2 \big) \underline{j} math
 * but at t = 0, object is at the origin **
 * so d = 0, so **


 * c) Find the time taken to reach the maximum height. **

math \\ . \qquad 30-gt=0 \\ .\\ . \qquad t \approx 3.1 \; \text{seconds} math
 * {Maximum height occurs when j component (vertical) of velocity is zero} **


 * d) Find the horizontal distance the stone travels before it lands. **

math \\ . \qquad 30t-\frac{1}{2}gt^2=0 \\ .\\ . \qquad t(30-\frac{1}{2}gt)=0 \\ .\\ . \qquad t=0 \; \text{ or } \; t \approx 6.1 math
 * {The stone hits the ground when the j (vertical) component of position is zero} **

math . \qquad \underline{r}(6.1)=245\underline{i}+0\underline{j} math
 * ignore t=0 because that was when stone was first projected. **


 * SO object hits ground approximately 245m from where it started.**

**See also:** www.mathsonline.com.au Y12Extension --> Calculus --> Physical Applications --> Lessons 5, 6

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