# A cannon shoots a projectile at 24 degrees from the horizontal. It lands on level ground 3000m down range. a) Find the initial velocity b) Calculate its maximum height?

Feb 20, 2017

$\textsf{\left(a\right)}$

$\textsf{199 \textcolor{w h i t e}{x} \text{m/s}}$

$\textsf{\left(b\right)}$

$\textsf{333.9 \textcolor{w h i t e}{x} m}$

#### Explanation:

Treat the vertical and horizontal components separately and then eliminate the time of flight.

$\textsf{\left(a\right)}$

The horizontal component of the velocity is constant so we can write:

$\textsf{v \cos \theta = \frac{d}{t}}$

$\therefore$$\textsf{t = \frac{d}{v \cos \theta} \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{t}$ is the time of flight

$\textsf{d}$ is the horizontal range

For the vertical component we can use the general equation:

$\textsf{v = u + a t}$

This becomes:

$\textsf{0 = v \sin \theta - \frac{\text{g} t}{2}}$

Note we divide the total time of flight by 2 because the time to reach maximum height (what we want) is the total time of flight / 2.

$\therefore$sf(vsintheta=("g"t)/2" "color(red)((2)))

Now we can substitute the value of $\textsf{t}$ from sf(color(red)((1)) into $\textsf{\textcolor{red}{\left(2\right)} \Rightarrow}$

$\textsf{v \sin \theta = \frac{g d}{2 v \cos \theta}}$

$\therefore$$\textsf{g d = 2 {v}^{2} \left(\sin \theta \cos \theta\right)}$

We can simplify this using the trig identity:

$\textsf{\sin 2 \theta = 2 \sin \theta \cos \theta}$

$\therefore$$\textsf{g d = {v}^{2} \sin 2 \theta}$

This gives $\textsf{d = \frac{{v}^{2} \sin 2 \theta}{g}}$ and is a useful formula which you can use to get the range.

However, we need to get v :

$\textsf{{v}^{2} = \frac{\mathrm{dg}}{\sin 2 \theta}}$

Putting in the numbers:

$\textsf{{v}^{2} = \frac{9.81 \times 3000}{\sin 48} = \frac{9.81 \times 3000}{0.7431} = 39604.36}$

$\therefore$$\textsf{v = \sqrt{39604.36} = 199 \textcolor{w h i t e}{x} \text{m/s}}$

$\textsf{\left(b\right)}$

We can use:

$\textsf{{v}^{2} = {u}^{2} + 2 a s}$

This becomes:

$\textsf{0 = {\left(v \sin \theta\right)}^{2} - 2 g h}$

$\therefore$$\textsf{h = {\left(v \sin \theta\right)}^{2} / \left(2 g\right)}$

$\therefore$$\textsf{h = {\left(199 \times 0.4067\right)}^{2} / \left(2 \times 9.81\right) = 333.9 \textcolor{w h i t e}{x} m}$

Feb 20, 2017

a) The initial velocity is 199 m/s; b) The maximum height is 334 m.

#### Explanation:

The general path of a projectile is shown below.

We can derive an equation for the initial velocity $v$ of the cannonball given the range $\text{R}$ and the angle of elevation θ.

For motion in the horizontal direction,

v_x = vcosθ

${v}_{x}$ is constant, so at any time the $x$ position of the cannonball is

x = v_xt = vtcosθ

For motion in the vertical direction,

v_y = vsinθ

${v}_{y}$ changes with time, because the cannonball is acted on by gravity. Thus,

v_y = vsinθ - g·t

At maximum height,

${v}_{y} = 0$

0 = vsinθ - g·t

The time to reach the top of the parabola is

t_"top" = (vsinθ)/g

It will take the same time for the cannonball to fall, so the total time in the air is

t = (2vsinθ)/g

Knowing the time allows us to find the range of the cannon:

R = v_xt = vcosθ × (2vsinθ)/g = (2v^2cosθsinθ)/g = (v^2sin2θ)/g

Rearranging gives

v = sqrt((gR)/(sin2θ))

$g = \text{9.81 m·s"^"-2}$
$R = \text{3000 m}$
θ = 24°

v = sqrt(("9.81 m·s"^"-2" × "3000 m")/sin(48°)) = sqrt("39 600 m"^2"s"^"-2") = "199 m·s"^"-1"

b) Maximum height

t_"top" = (vsinθ)/g

The distance traveled by a falling object is

h = 1/2g·t^2

Therefore, the maximum height is

h = 1/2g((vsinθ)/g)^2 = (v^2sin^2θ)/(2g) = ((199 "m"·color(red)(cancel(color(black)("s"^"-1"))))^2 × sin^2(24°))/(2 × 9.81 "m"·color(red)(cancel(color(black)("s"^"-2"))))

$h = \frac{6551}{19.62} \textcolor{w h i t e}{l} \text{m" = "334 m}$