Projectile question (basic level)

[Suyash Singh]

Homework Statement

The maximum range of a bullet fired from a toy pistol mounted on a car at rest is R0=40m .
What will be the acute angle of inclination of the pistol for maximum range when the car is moving in the direction of firing with uniform v=velocity 20m/s, on a horizontal surface ? (g=10m/s).

Homework Equations

T=2 u sin(theta)/g
r=u u sin2(theta)/g
h= u u sin(theta) sin(theta)/2g

The Attempt at a Solution

Maximum range is always at 45 degrees so no calculation needed.( am i right?)

different answers are given everywhere for this same question.

 

[kuruman]

Maximum range is always at 45 degrees so no calculation needed.( am i right?)

You are not right. The 45° is the maximum range angle when the velocity of the pistol is zero relative to the ground. Here the pistol is moving.

 

[Suyash Singh]

You are not right. The 45° is the maximum range angle when the velocity of the pistol is zero relative to the ground. Here the pistol is moving.

how do i do this question? Everywhere there is a different or incomplete solution.

 

[kuruman]

First you need to find the velocity of the bullet relative to the pistol.

 

[Suyash Singh]

First you need to find the velocity of the bullet relative to the pistol.

u= 20 m/s

 

[kuruman]

u= 20 m/s

That's the velocity of the car relative to the ground. Please read the problem carefully and try again. You want to find how fast the projectile is moving away from the pistol regardless of whether the car is moving or not.

On Edit: Sorry, I just realized that the velocity of the projectile relative to the pistol is also 20 m/s. OK, can you write an expression for the range in terms of the x and y components of the projectile's velocity relative to the ground?

 

[Suyash Singh]

y=x tan(theta) - g x x/2 u u cos(theta) cos(theta)

 

[kuruman]

This is the equation of the parabola y(x) that the projectile describes in space when the pistol is not moving. This is not the range. What is an expression for the range in terms of the x and y components of the initial velocity? Your expression must contain terms v_0x and v_0y.

 

[Suyash Singh]

r= u^2 sin2(theta)/g
where u = square root(v0x^2 + v0y^2)

 

[kuruman]

Right. Here θ is the angle from the horizontal; v_0x and v_0y are the components of the initial velocity relative to the ground. What are these components when the projectile is shot from a car that is moving at 20 m/s relative to the ground?

 

[Suyash Singh]

i don't know how to find that.

But my guess is that Vox=20m/s
i don't know about Voy.

 

[kuruman]

But my guess is that Vox=20m/s

Not a very good guess. That would be the the case when the car is not moving. Suppose you are in the car shooting the pistol and your friend is standing on the road watching. You see the projectile move away from you in the horizontal direction at 20 m/s. Your friend sees you, the pistol and the car move at 20 m/s horizontally. Do you think he would see the projectile also move at 20 m/s in the horizontal direction? Please try to think this through again. We'll worry about v_0y after you have figured this one out.

 

[gmax137]

i don't know how to find that.

Maybe I am making this problem more complicated than it needs to be, but here's what I did.

First, make a sketch, showing the velocity vector of the car, and the velocity vector of the bullet relative to the car. I could then write an expression for the sum of the two vectors in terms of the other vectors. This expression will have the (unknown) included angle (theta). That's the first step in my solution. Then I break that into its X and Y components.

 

[haruspex]

That would be the the case when the car is not moving.

Did you mean, that would be the horizontal velocity if the gun were aimed straight up?

 

[kuruman]

Did you mean, that would be the horizontal velocity if the gun were aimed straight up?

I meant that 20 m/s would be the velocity if the pistol were fired at zero elevation angle from a motionless car, the tacit inference being that the horizontal component must be greater than 20 m/s. It's also, of course, the horizontal velocity when the pistol is fired straight up from the moving car, but I am not sure that OP is there yet.

 

[Suyash Singh]

Vox = speed of car in horizontal direction + speed of bullet with respect to gun in horizontal direction
= 40 m/s

Similarly,
Voy= 20 m/s (only bullet velocity now)

Did i get it right?

 

[kuruman]

Did i get it right?

You got it right but only if the pistol is fired horizontally. Here the pistol is fired at an angle, call it θ, and you are looking for a value of θ such that the range is maximum. So you need to find an expression for the range in terms of θ and maximize that. To do that, first you need to find v_0x and v_0y relative to the ground when the pistol is fired at angle θ. Relative to the car, the initial velocity components are {v₀cosθ, v₀sinθ} where v₀ is the speed of the projectile relative to the pistol. What do you think they are relative to the ground if the pistol is moving horizontally with speed u?

 

[Vriska]

do you think the bullet fired by the gun will make the same angle with respect to an external observer as the one in the car?

 

[Suyash Singh]

do you think the bullet fired by the gun will make the same angle with respect to an external observer as the one in the car?

yes

 

[Suyash Singh]

You got it right but only if the pistol is fired horizontally. Here the pistol is fired at an angle, call it θ, and you are looking for a value of θ such that the range is maximum. So you need to find an expression for the range in terms of θ and maximize that. To do that, first you need to find v_0x and v_0y relative to the ground when the pistol is fired at angle θ. Relative to the car, the initial velocity components are {v₀cosθ, v₀sinθ} where v₀ is the speed of the projectile relative to the pistol. What do you think they are relative to the ground if the pistol is moving horizontally with speed u?

Ok
Vox= 20cos(theta)+20
Voy=20sin(theta)

 

[kuruman]

Vox= 20cos(theta)+20
Voy=20sin(theta)

Very good. Remember that the angle θ in these expressions is the angle measured by you riding on the car. Now, having these components, can you find the angle of projection relative to your friend standing at rest on the ground as @Vriska suggested? Is it also θ or something else? Hint: It's something else.

 

[Vriska]

yes

So if I throw a ball vertically up from a moving car, will an observe from the outside see it as moving vertically? if they do see it vertically from ground , wouldn't the ball land on the place immediately below where it was thrown? does this happen?

 

[haruspex]

So if I throw a ball vertically up from a moving car, will an observe from the outside see it as moving vertically?

It depends what that condition means. To clarify, I believe you mean that it looks vertical to you in the car.

 

[kuruman]

t depends what that condition means. If you mean that it looks vertical to you in the car then no, the observer in the ground frame will not see it as vertical. The car's horizontal velocity will be added.

I believe @Vriska was responding to OP's post #19 in an attempt to show the logical conclusion of OP's answer.

 

[haruspex]

I believe @Vriska was responding to OP's post #19 in an attempt to show the logical conclusion of OP's answer.

Thanks - edited.

 

[Suyash Singh]

Very good. Remember that the angle θ in these expressions is the angle measured by you riding on the car. Now, having these components, can you find the angle of projection relative to your friend standing at rest on the ground as @Vriska suggested? Is it also θ or something else? Hint: It's something else.

Sorry internet was very slow yesterday.
I can't understand why do we need to find the angle with respect to friend on ground at rest.Also i am not able to figure out how to find it.

something like this i think?
angle of projectile wrt car=angle of projectile wrt ground - angle of car wrt ground
where wrt means 'with respect to'

 

[Suyash Singh]

So if I throw a ball vertically up from a moving car, will an observe from the outside see it as moving vertically? if they do see it vertically from ground , wouldn't the ball land on the place immediately below where it was thrown? does this happen?

yes it does

 

[kuruman]

yes it does

Please watch this video. It should clear up this point for you.

I can't understand why do we need to find the angle with respect to friend on ground at rest.Also i am not able to figure out how to find it.

OK, we will go ahead and finish the problem without the projection angle with respect to the friend on the ground. Can you derive or just write down an expression for the range that involves only the x and y components of the velocity but not the angle of projection? Hint: sin(2θ) = 2 sinθ cosθ.

 

[haruspex]

yes it does

As I indicated in post #23 there was a small ambiguity in Vriska's question.
Vriska means: if it looks to you in the cart as though it is going straight up above you then what will it look like to a bystander on the ground?

 

[Suyash Singh]

omg the video was amazing. My brain is rejecting this information i don't know why.

R=u u sin(2θ)/g = u u 2 sinθ cosθ/g = (u u 2 voy/v vox/v)/g
= (u u 2 Vox Voy)/(g v v)
where vox and voy are x and y components of velocity wrt car.

 

[Suyash Singh]

As I indicated in post #23 there was a small ambiguity in Vriska's question.
Vriska means: if it looks to you in the cart as though it is going straight up above you then what will it look like to a bystander on the ground?

ok now i get what he meant by watching that video in #28

 

[kuruman]

R=u u sin(2θ)/g = u u 2 sinθ cosθ/g = (u u 2 voy/v vox/v)/g
= (u u 2 Vox Voy)/(g v v)
where vox and voy are x and y components of velocity wrt car.

No. For any projectile fired with initial speed $v_0$ relative to the ground, $$R=\frac{v_0^2\sin(2\theta)}{g}=\frac{2(v_0\sin \theta)(v_0 \cos\theta)}{g}=\frac{2v_{0y}v_{0x}}{g}$$
This equation gives the range if the projectile returns to the same vertical height from which it was fired. Now replace $v_{0x}$ and $v_{0y}$ with the expressions you found in post #20 and find $\theta$ such that $R$ is maximum.

 

[Suyash Singh]

R=2v0yv0x / g
= 2 (20cos(theta)+20)(20sin(theta))/g

now i use differentiation i suppose? to get 60 degrees

 

[kuruman]

now i use differentiation i suppose? to get 60 degrees

Yes. What expression did you get when you differentiated?

 

[Suyash Singh]

2cos(2theta)=-cos(theta)

2(cos^2θ-sin^2θ)=-cos(θ)

2(2cos^2θ-1)=-cos(theta)

then i solve it like a quadratic equation and then i differentiate it again to find maxima and minima