Relative velocity of a ball thrown inside a moving truck

[jisbon]

Homework Statement

A truck travels at a speed v to the right (relative to the ground). A ball
launched inside the truck at tan theta = 4/3 has the same initial speed v (relative to truck).
An observer inside the truck measures the ball lands at a distance L from the initial
point. At the same time another observer on the ground measures the ball travel a
distance L' = xL. Determine x.

Relevant Equations

vx = vcos theta
vy = vsin theta

Hi all,

Not sure on how to start this question in the first place, but from what I gathered from the data given

I managed to derive this from the question:

$\theta = 53.13\deg$

Let inside truck be t, final position be f, and ground be g

$D_{ft} = L$
$D_{fg} = xL$

Also, for velocity (I'm pretty sure I'm wrong here)
Let velocity of ball be b.

$Vx_{bt} = cos \theta v$
$Vx_{bg} = v$

Not sure how to start this question, should I be using any form of kinematics equation or sort? Thanks

 

[BvU]

You had a nice $R = v^2 \sin(2\theta)/g \$ equation in your other thread.
What are these factors for the driver ? and for the outside observer ?

By writing them out for each of the two you can divide away a bunch of them in $\ \ R_{\rm driver} / R_{observer} = x \ \$!

 

[jisbon]

=

You had a nice $R = v^2 \sin(2\theta)/g \$ equation in your other thread.
What are these factors for the driver ? and for the outside observer ?

By writing them out for each of the two you can divide away a bunch of them in $\ \ R_{\rm driver} / R_{observer} = x \ \$!

Hi.
Thanks so much for helping out.
Here's what I thought about this question:
The observer inside the truck that is measuring distance can only see the velocity from the ball itself since he is also in the truck moving, hence
$R = v^2 \sin(2\theta)/g \$
$R = v^2 \sin(0.96)/9.8 \$
However, observer at ground can see both velocity from truck and ball, hence ball 'inherit' velocity from truck too. In that case, how do I include the horizontal velocity component from the truck into this formula: $R = v^2 \sin(2\theta)/g \$ ?
Thanks

 

[BvU]

Make a good sketch ! Showing the horizontal and the vertical components.

 

[jisbon]

Make a good sketch ! Showing the horizontal and the vertical components.

Will observer be seeing ball as having horizontal velocity of v + vsin theta?

 

[PeroK]

View attachment 248427
Will observer be seeing ball as having horizontal velocity of v + vsin theta?

Yes. Depending on your choice of theta. It's more usual to take theta to the horizontal.

 

[BvU]

It's more usual to take theta to the horizontal.

As @jisbon did in post #1. I suppose it's a mistake.

So in the full expression for $\ \ R_{\rm driver} / R_{observer} = x \ \$, there is one $g$ and there are two each of $v$ and $\theta$.

Hurray ! The $g$ divides out. If the driver assigns the unprimed symbols and the external observer the primed symbols, what can you say about $v'$ and $\theta'$ ?

 

[PeroK]

As @jisbon did in post #1. I suppose it's a mistake.

So in the full expression for $\ \ R_{\rm driver} / R_{observer} = x \ \$, there is one $g$ and there are two each of $v$ and $\theta$.

Hurray ! The $g$ divides out. If the driver assigns the unprimed symbols and the external observer the primed symbols, what can you say about $v'$ and $\theta'$ ?

There's no need to calculate $\theta'$, unless you want to make things complicated!

 

[BvU]

That's going to save us a number of posts

But it sure is a good hint for @jisbon : after all you are looking for the ratio $\sin(2\theta)\over \sin(2\theta')$ !

 

[PeroK]

That's going to save us a number of posts

But it sure is a good hint for @jisbon : after all you are looking for the ratio $\sin(2\theta)\over \sin(2\theta')$ !

My hint is that $t' = t$.

 

[jisbon]

Um haha not sure what are y'all discussing on. But basing on the fact that observer will be seeing ball as having horizontal velocity of v + vsin theta

Is it true to assume that

$\frac{V^2sin2\theta}{g}÷\frac{(V^2sin^2\theta+(V^2+V^2cos^2\theta)) sin2\theta }{g} = x$

whereby

$(V^2sin^2\theta+(V^2+V^2cos^2\theta)$ is the velocity seen by the observer?

 

[kuruman]

Stop and think for the moment.
1. If the truck observer sees a horizontal component $V_{0x}$, what is the horizontal component $V'_{0x}$ as seen by the ground observer?
2. If the truck observer sees a vertical component $V_{0y}$, what is the vertical component $V'_{0y}$ as seen by the ground observer?
First put the two components together to get the velocity as seen by the ground observer, then find the range as seen by the ground observer and finally take the ratio to find $x$.

I do not mean to divert you from the course that @BvU and @PeroK have charted for you, but it is worth noting that an equivalent expression for the horizontal range is $R=\dfrac{2V_{0x}V_{0y}}{g}$ and the velocity components as seen by the truck observer are the right sides of a 3-4-5 triangle. If you answer the two questions above, you should be able to find the ratio without having to worry about $\theta$.

 

[jisbon]

1. horizontal component as seen by the ground observer is more than horizontal component seen by truck observer by V.

2. Both observer sees the same vertical component.

Will this be correct?

 

[PeroK]

1. horizontal component as seen by the ground observer is more than horizontal component seen by truck observer by V.

2. Both observer sees the same vertical component.

Will this be correct?

Yes. Point 1. is true because the truck is moving. The ground observer sees the additional motion of the truck.

What does point 2 imply? Hint gravity and time.

 

[jisbon]

As per all your help, I have gotten the accurate answer. The only problem/question I'm facing is:

an equivalent expression for the horizontal range is $\dfrac{2V_{0x}V_{0y}}{g}$

how did you manage to get this expression? Thanks :)

 

[BvU]

It is your own equation in the other thread (The $t_1-t_2$ one) !

 

[kuruman]

As per all your help, I have gotten the accurate answer. The only problem/question I'm facing is:how did you manage to get this expression? Thanks :)

$$\sin(2\theta)=2\sin\theta\cos\theta~~~\mathrm{(trig~identity)}$$ $$V_0^2 \sin(2\theta )=2(V_0 \sin\theta)( V_0 \cos\theta)=2V_{0x} V_{0y}$$ $$R=\frac{V_0^2 \sin(2\theta )}{g}=\frac{2V_{0x} V_{0y}}{g}.$$

 

[PeroK]

$$\sin(2\theta)=2\sin\theta\cos\theta~~~\mathrm{(trig~identity)}$$ $$V_0^2 \sin(2\theta )=2(V_0 \sin\theta)( V_0 \cos\theta)=2V_{0x} V_{0y}$$ $$R=\frac{V_0^2 \sin(2\theta )}{g}=\frac{2V_{0x} V_{0y}}{g}.$$

Or:

$R = V_x t$

And $t = \frac{2V_{0y}}{g}$

Where $t$ is the time of flight.

 

[BvU]

Homework Statement: When 2 projectile motions are launched with velocity v, one at angle x and the other one at angle 90-x Derive an expression for the time difference in terms of range, x and g
Homework Equations: R = v^2 sin2x / g

 

[jisbon]

Oh thanks :) only thought of this : R = v^2 sin2x / g