Projectile's Maximum Speed

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One of the practice physics problems I was working on involves this scenario: "A gun shoots a projectile at some angle theta from the horizontal. The gun that shoots it is a certain height above the ground." One question asks when the maximum velocity of the projectile is reached. I narrowed it down to either right before the projectile hits the ground, or right as it is launched from the gun. The book, without explaining, says that the maximum velocity is reached right before it hits the ground (all it says is 'because the gun is above the ground'). Is there a reason for this that someone could explain, or is it more likely that I'm missing something in the problem? Thanks

 

[cge0]

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[dmission]

Vector component x and y are maximized at that point. It has achieved maximal vertical velocity and the horizontal velocity is constant (ignoring air resistance).

Wouldn't vector component y be the same everywhere? How do you know, though, that the X component is a larger speed there than it was when it was shot from the gun (at some angle, theta)?

 

[jissho]

Hi, the key here is that the gun is above the ground. When the bullet is initially shot form the gun it has velocity a. As the bullet rises in the air it will reach vmax and then start to fall. Eventually the bullet will reach the height that it was originally shot from and have velocity -a. Since the bullet was shot above the ground it will continue to fall and pick up speed after it reaches the height that it was shot from. This means that the magnitude of the velocity (speed) when it hits the ground will be larger than the magnitude of the initial velocity.

 

[paul411]

Wouldn't vector component y be the same everywhere? How do you know, though, that the X component is a larger speed there than it was when it was shot from the gun (at some angle, theta)?

The x component remains the same throughout because there is no external force acting on it.

The y component changes because gravity acts on it.

 

[GyroGyro]

I think the above posters already explained it well, but just in case,

Vx should not change throughout the flight so don't worry about it.

Vy DOES change since it is acted upon by an acceleration (gravity that is). Think about the launch of a regular projectile. When that projectile is moving up to its maximum height, the Vy is continually decreasing to 0. Then, when it goes down from maximum height, Vy should be decreasing. When it reaches the exact value of when it was launched at (i.e. no vertical displacement), it has the exact same Vy but in the opposite direction.

Using the equation (if this helps), Vy (final) ^2 = Vy (initial) ^2 + 2 (a) (x). In this case, x is equal to 0 when it is at the level of the gun.

Thus Vy (final) ^2 = Vy (initial) ^2 (the magnitdues of the Vy are the same, but the directions are opposite).

If it falls PAST the gun (note, a situation like this is almost always ignored to be honest as the height of the gun would be considered negligible), it continues to accelerate downwards and Vy gets even lower (use the above equation if you its confusing). Thus, the magnitude of the velocity should be higher when it is about to hit the ground.

 

[dmission]

Thanks for all the replies. So, just to make sure I understand: so if the gun were not raised off the ground, then right before the bullet hits the ground at the end of its path, it would have around the same Vy?

 

[GyroGyro]

Yes, the magnitude would be the same, but the direction would be opposite (so if Vy had a value of 3 m/s when launched, then right when it hits the ground (assuming the gun has negligible height and it falls on even ground (in other words total y displacement = 0) then Vy would have a value of -3 m/s). Make sure you remember that velocity has a magnitude and direction component whereas speed does not (speed does not have direction).

To be honest, I think the problem is poorly worded in your book. As you have written it, the question says when is the maximum velocity reached. It should say magnitude somewhere in there because a negative number is less than a positive number.. Actually maybe my line of thinking here is wrong. I just don't like that wording.

 

[dmission]

Ah you're right, it does say speed. My mistake; thanks for the clarification

 

[neurodoc]

This problem assumes that the projectile leaves the gun with a certain constant horizontal velocity unaffected by air resistance, etc. Therefore the only force acting on the projectile is gravity in a downward vertical direction.

That's not a real-life situation, but it is what the problem states.