# Relative velocity of a ball thrown inside a moving truck

• jisbon
In summary, the conversation discusses the calculation of the horizontal range of a ball thrown from a truck. It involves finding the ratio between the horizontal range as seen by the driver inside the truck and the horizontal range as seen by an observer outside the truck. The discussion also includes determining the horizontal and vertical components of velocity as seen by the ground observer and using them to calculate the range. The conversation also mentions an equivalent expression for the horizontal range and the velocity components as seen by the truck observer. Finally, the conversation provides a hint for finding the ratio without involving the angle theta.
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

Last edited by a moderator:
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 \ \ ##!

=
BvU said:
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.
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

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

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

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

jisbon said:
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.

PeroK said:
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'## ?

BvU said:
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!

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') ## !

BvU said:
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##.

BvU
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?

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##.

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?

jisbon said:
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.

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

kuruman said:
an equivalent expression for the horizontal range is ##\dfrac{2V_{0x}V_{0y}}{g}##
how did you manage to get this expression? Thanks :)

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

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

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

##R = V_x t##

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

Where ##t## is the time of flight.

jisbon said:
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

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

## 1. What is relative velocity?

Relative velocity is the measurement of the velocity (speed and direction) of an object in relation to another object. It takes into account the motion of both objects and can be described as how fast one object is moving with respect to the other.

## 2. How is relative velocity calculated?

Relative velocity can be calculated by subtracting the velocity of one object from the velocity of the other object, taking into account their respective directions. This is known as vector subtraction. Alternatively, it can also be calculated using the formula vAB = vA - vB, where vAB is the relative velocity of A with respect to B, vA is the velocity of object A, and vB is the velocity of object B.

## 3. How does the velocity of a ball thrown inside a moving truck differ from the velocity of a ball thrown on the ground?

The velocity of a ball thrown inside a moving truck is affected by both the initial velocity of the throw and the velocity of the truck. This means that the ball's velocity will be a combination of these two velocities, resulting in a different velocity than if the ball were thrown on the ground.

## 4. What factors can affect the relative velocity of a ball thrown inside a moving truck?

The relative velocity of a ball thrown inside a moving truck can be affected by the initial velocity of the throw, the velocity of the truck, and any external forces acting on the ball such as air resistance or friction.

## 5. How does the direction of the ball's throw impact its relative velocity inside a moving truck?

The direction of the ball's throw will impact its relative velocity inside a moving truck as it will determine the angle at which the ball is moving in relation to the truck's velocity. This can result in a change in the overall velocity of the ball and its direction within the truck.

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