# Math Help - kinematics motion problem

1. ## kinematics motion problem

A football kicks a ball of a cliff overlooking the ocean, the cliff 11.2m above sea level. And she kicks it with initial speed $10ms^{-1}$ at an angle 30 degrees measured upwards from the horizontal .

Take 9 = $10ms^{-2}$

1) Assuming that there is no air resistance, how far from the cliff in the horizontal direction

does it hit the ocean?

2) Now suppose that the air resistance can be modelled by a force $f = -kv$ , with k = 0.200kgm^{-1} . Assume that the mass of the football iss 0.2kg.

2a) Write down the equations of motion for the vertical and the horizontal directions.

2b) For motion in the vertical direction, integrate the equation of motion twice to find
a formula for the height of the football as a function of time.

for part 1 I am not sure how to work out the distance, I have resolved the forces into its components

$10m^{-1} \times cos 30 = 10 \times 1/2 = 5$

$10 m^{-1} \times sin 30$

but how do I work out distance?

and really not sure how to start with on 2a,2b.

any help appreciated.

2. ## Re: kinematics motion problem

For number one, gravity is only working in the y-direction. After the ball is kicked, it has an original velocity in both the x- and y- directions, but after that, it is simply acted on by gravity in the y-direction. So, you calculated the initial velocity in the x- and y- directions. Since there are no forces acting on the ball in the x- direction, the equation of the ball's horizontal distance is $x(t) = v_x t = 5t$. In the y-direction, the ball started 11.2m up. Initial velocity is $5\sqrt{3}$. Gravity has an acceleration of $-10$ (I am ignoring stripping units as I prepare to plug this into an equation). So, $y(t) = y_0 + v_y t + a_y t^2 = 11.2 + 5\sqrt{3} t - 10t^2$. You want the value of $t$ when the ball hits the water (sea-level), so you want $y(t)=0$. That is a quadratic equation. Once you know the value of $t$, plug it into $x(t)$ to determine the ball's horizontal position at the time it hit the water.

For 2, you have a force that depends on velocity. In the horizontal direction, the acceleration is $a_x(t) = -0.2v_x(t)$. In the vertical direction, acceleration is $a_y(t) = -10 - 0.2v_y(t)$. That should get you started.

3. ## Re: kinematics motion problem

Okay thank you, I understand part 1 now, but how did you know the equation $a_{x}(t) = -0.2v_{x}(t)$ ?

4. ## Re: kinematics motion problem

Originally Posted by Tweety
Okay thank you, I understand part 1 now, but how did you know the equation $a_{x}(t) = -0.2v_{x}(t)$ ?
Sorry. I was wrong. For part (1) I mixed up the initial x- and y-velocities. Swap the $5$ and $5\sqrt{3}$. For (2), I forgot to divide out the mass. Force is mass times accelerations, so $-0.2v = 0.2a$ is the formula. That means $a_x(t) = -v_x(t)$ and $a_y(t)=-10-v_y(t)$. This gives two differential equations. Solving them, you get:

$v_x(t) = c_1e^{-t}$ and $v_y(t) = c_2e^{-t}-10$

Plugging in initial values: $v_x(0) = 5\sqrt{3} = c_1 e^{-0} = c_1$, so $v_x(t) = 5\sqrt{3} e^{-t}$.
$v_y(0) = 5 = c_2e^{-0}-10 = c_2-10$, so $c_2 = 15$. This gives $v_y(t) = 15e^{-t}-10$.

Then, integrating each, you get $s_x(t) = -5\sqrt{3} e^{-t} + c_3$ and $s_y(t) = -15e^{-t}-10t+c_4$

Plugging in initial values:

$s_x(0) = 0 = -5\sqrt{3}e^{-0}+c_3 = -5\sqrt{3}+c_3$, so $c_3 = 5\sqrt{3}$
$s_x(t) = 5\sqrt{3}(1-e^{-t})$

$s_y(0) = 11.2 = -15e^{-0} - 10(0) + c_4 = -15+c_4$, so $c_4=26.2$.
$s_y(t) = 26.2-15e^{-t}-10t$

5. ## Re: kinematics motion problem

Hmm, the force for air resistance has the wrong units. Should it be $f = -kv^2$? That would give the correct units. Then $a_x(t) = -(v_x(t))^2$ and $a_y(t) = -10-(v_y(t))^2$. This gives $v_x(t) = \dfrac{1}{c_1+t}$ and $v_y(t) = -\sqrt{10} \tan\left(\sqrt{10}(c_2 + t)\right)$.

This gets messy really quickly, so I doubt it is what your professor is looking for, but still, the units are not correct otherwise.

6. ## Re: kinematics motion problem

If you do want to solve the problem when air resistance is $f = -kv^2$, then:

$v_x(0) = 5\sqrt{3} = \dfrac{1}{c_1+0}$, so $c_1 = \dfrac{1}{5\sqrt{3}}$
$v_x(t) = \dfrac{5\sqrt{3}}{1+5\sqrt{3}t}$

$v_y(0) = 5 = -\sqrt{10}\tan\left(\sqrt{10}(c_2+0)\right)$, so $c_2 = \dfrac{\pi n - \arctan\left(\sqrt{\dfrac{5}{2}}\right) }{\sqrt{10} }, n \in \mathbb{Z}$
$v_y(t) = -\sqrt{10}\tan\left(\sqrt{10}t - n\pi - \arctan\left(\sqrt{2.5}\right) \right)$

Integrating each, we get:
$s_x(t) = \ln\left|5\sqrt{3} t + 1\right| + c_3$
$s_y(t) = \ln\left| \cos\left(n\pi + \arctan(\sqrt{2.5}) - \sqrt{10}t\right) \right|+c_4$

Using the sum of angles formula, we have

\displaystyle \begin{align*} \cos(n\pi + \arctan(\sqrt{2.5}) - \sqrt{10}t) & = \cos( n\pi ) \cos( \arctan(\sqrt{2.5}) - \sqrt{10}t ) - \sin( n\pi ) \sin( \arctan(\sqrt{2.5})-\sqrt{10}t) \\ & = (-1)^n\cos(\arctan(\sqrt{2.5}) - \sqrt{10}t) \end{align*}

So, the choice of $n$ is arbitrary since we are taking the absolute value. Hence, we can assume $n=0$.

Again, plugging in initial values:
$s_x(0) = 0 = \ln\left|5\sqrt{3}\cdot 0 + 1\right| + c_3 = c_3$, so
$s_x(t) = \ln\left|5\sqrt{3}t + 1\right|$

$s_y(0) = 11.2 = \ln \left| \cos\left(\arctan(\sqrt{2.5}) - \sqrt{10}\cdot 0\right) \right| + c_4 = \ln \left|\dfrac{2}{\sqrt{14}} \right|+c_4$, so $c_4 = 11.2 + \ln\left|\dfrac{\sqrt{14}}{2}\right|$
$s_y(t) = 11.2 + \ln\left|\dfrac{\sqrt{14}}{2}\right| + \ln \left| \cos\left(\arctan(\sqrt{2.5}) - \sqrt{10}t\right) \right|$