Problem Solving Strategies

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John Rennie

04:28

@Pizza Thanks :-)

@BinkyMcSquigglebottom That's solving for 𝑡 using the height change then using x = vt

5 hours later…

Swan

09:09

@JohnRennie I have seen the solutions to this problem but I wanted to know if we can solve this question using a phasor diagram? As in, can we draw phasor diagram for when the 2 forces acting on a partile aren't linear?

John Rennie

Hi :-)

I don't think this would work. The trouble is that we don't have a linear system here i.e. you can't just add the two displacements.

I would do it by splitting s(t) into x and y components, and then adding the x component to the first motion, which I'm guessing is the solution you have seen already.

Swan

09:27

@JohnRennie Thanks :-)

Pizza

10:09

@JohnRennie for Binky's exercise, that's what you mean, right?

$v_{0x}=v_0\cdot\cos(\theta)$
$v_{0y}=v_0\cdot\sin(\theta)$
$d_x =v_{0x}\cdot t \Rightarrow d_x =v_0\cdot\cos(\theta)\cdot t \Rightarrow t=\frac{d_x}{v_0\cdot\cos(\theta)}$ $\quad(1)$
$d_y = v_{0y}\cdot t-\frac{1}{2}gt^2\Rightarrow d_y=v_0\cdot\sin(\theta) \cdot \frac{d}{v_0\cdot\cos(\theta)} - \frac{1}{2}g\cdot\left(\frac{d}{v_0\cdot\cos(\theta)}\right)^2$ $\quad(2)$$$t=\frac{9400m}{v_0\cdot\cos(35°)}\tag{1}$$$$-3300m=v_0\cdot\sin(35°)\cdot\frac{9400m}{v_0\cdot\cos(35°)}- \frac{1}{2}g\left(\frac{9400m}{v_0\cdot\cos(35°)}\right)^2\tag{2}$$

John Rennie

@Pizza Yes

There are several approaches to the problem, but they all basically involve x(t) = 9400 and y(t) = -3300 at the same time 𝑡.

Pizza

10:41

$$-3300m = 9400m \cdot \tan(35°) - \frac{1}{2} 9.81m/s^2\cdot\frac{(9400m)^2}{v_0^2 \cos^2(35°)}$$$$9881m = 4,905m/s^2\cdot\frac{88360000m^2}{v^2_0 \cdot 0,671}$$$$v^2_0\cdot9881m=4,905m/s^2\cdot \frac{88360000m^2}{0,671}$$$$v^2_0=4,905m/s^2\cdot\frac{88360000m^2}{9881m\cdot0,671}=\frac{433405800m^3/s^2}{6630,151m}\approx65368,9m^2/s^2$$$$v_0=\sqrt{65368,9m^2/s^2}\approx 255m/s$$

John Rennie

Yes, that's the answer I got.

Pizza

if we wanted to find the flight time, we can do it like this, right?

$$t_f=\frac{d_x}{v_0\cos(\theta)}=\frac{9400m}{255m/s\cdot\cos(35°)}=45s$$

where $t_f=$ flight time and $d_x =$ the distance on the $x$ axis

@JohnRennie however, without a calculator the calculation of the first point was a little complicated 😂

Binky McSquigglebottom

@JohnRennie but why if I try to find $v_0$ from that formula do I find it different?

John Rennie

@Pizza Yes, I did it using Google Sheets. When I have involved calculations I often lay them out in a spreadsheet to keep then clear.

@BinkyMcSquigglebottom Hi :-)

Pizza

@JohnRennie nice idea!

hi @BinkyMcSquigglebottom

John Rennie

10:56

23 hours ago, by Binky McSquigglebottom

9400=x^2Sin[2*35°]/2*9,81(1+Sqrt[1+2*9,81*3300/x^2(sin(2*35°))^2)]

There is a possible error in the equation. I don't know exactly how Wolfram would parse it, but 2*9.81 should be in brackets.

9400=x^2Sin[2*35°]/(2*9,81) * (1+Sqrt[1+2*9,81*3300/x^2(sin(2*35°))^2)])

18 hours ago, by Pizza

$$x_G = \frac{v^2_0\sin 2\theta}{2g} \left( 1 + \sqrt{1 + \frac{2g \ y_0}{v^2_0 \sin^2 \theta} \ }\right)$$

This is the equation

Binky McSquigglebottom

the same is not found

😭

John Rennie

Also note that y₀ needs to be negative because we are taking the upwards direction to be positive so a height below the starting point is negative.

@BinkyMcSquigglebottom I'm not sure what that means ...

Binky McSquigglebottom

nothing changes

I still can't find the right solution

John Rennie

Do you want to go through how I did the calculation?

I'm only here for another 20 minutes. After that we'll have to continue tomorrow.

Pizza

11:13

$$9400 \text{ m} = \frac{v^2_0 \cdot 0.9397}{2 \cdot 9.8 \text{ m/s}^2} \left( 1 + \sqrt{1 + \frac{2 \cdot 9.8 \text{ m/s}^2 \cdot (-3300 \text{ m})}{v^2_0 \cdot \sin^2(35°)}} \right)$$

@JohnRennie it should look like this, using @BinkyMcSquigglebottom formula, right?

John Rennie

I guess so. I used a different method and I have not checked their equation. But you got the correct answer so it must be right :-)

Pizza

@JohnRennie I think he wants to use this equation

but he said "I still can't find the right solution"

he wasn't referring to your method (the one I used previously)

John Rennie

Ah, OK, you used the same method as me then?

Pizza

yes

the formula that I sent now is the @BinkyMcSquigglebottom one

John Rennie

I'd have to go through their equation to check it but I'm not at my desk so I'll have to do it later.

Pizza

11:19

In my opinion this is much more complicated, it is better to use your method

on wikipedia it says that the @BinkyMcSquigglebottom formula is: the case in which the starting height is not zero. All we need to do is reuse the function y(x) by adding the constant $y_0 \ne 0$.

John Rennie

I agree! :-)

Pizza

$$9400 \text{ m} = \frac{v^2_0 \cdot 0.9397}{19.6 \text{ m/s}^2} \left( 1 + \sqrt{1 - \frac{64680m^2/s^2}{v^2_0 \cdot 0,329}} \right)$$

there are 2 $v_0$ variables, I don't know how to get out of it

John Rennie

I guess that's why they used Wolfram.

Pizza

ye...

John Rennie

It's not that hard but it's messy.

Pizza