Projectile Motion angled downwards

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SUMMARY

The discussion focuses on calculating the projectile motion of a bullet fired at a downward angle of 40 degrees with a muzzle velocity of 350 m/s from a height of 600 m. Key calculations include determining the time to hit the ground, horizontal distance traveled, and final vertical velocity. The initial vertical velocity is calculated as -224.98 m/s, while the horizontal velocity remains constant at 268.11 m/s. The projectile motion remains parabolic despite the downward angle, as the trajectory can be represented as a segment of a parabolic curve.

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NYmike
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Hello all. I am currently taking an AP Physics class in high school. My teacher has given us a homework assignment involving projectile motion, however, it is different than anything else we have done. Usually, the angle is upward, or directly horizontal...This question is downward. My textbook does not offer any examples, so I am unsure of where to go for help.

A gun is pointed down at an angle 40 degress below the horizontal. The muzzle velocity is 350 m/s and the bullets mass is 30 g. The gun is 600 m above the ground. Calculate how long it takes to hit the ground, the horizontal distance traveled, the vertical velocity as it hits the ground, ...

The mass is useless, but it's given. There is a list of 10 or so questions, but I just need help getting started. I don't want 1 error to snowball.

In a normal parabola, the final vertical velocity is the negative value of the inital vertical velocity. However...this wouldn't be a normal parabola, right? Because it's just getting shot diagonally in a downward motion.

VERTICAL:

vi = 224.98 m/s
vf = -224.98 m/s
t =
d = -600 m
a = -9.8 m/s^2 (neglect air resistance)

HORIZONTAL:

vi = 268.11 m/s
vf = 268.11 m/s
t =
d =
a = 0 m/s^2


Would that be correct for the givens? I took sin and cos of 40. I am just not sure if the final vertical velocity is correct, because I don't think it would be a parabola.

Sorry for writing so much, and thank you for any help that you can provide =D
 
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Your horizontal information is correct (I did not check your arithmetic).But you vertical information is not. The initial velocity should be negative, you need to compute the final velocity with the information given.
 
Integral said:
Your horizontal information is correct (I did not check your arithmetic).But you vertical information is not. The initial velocity should be negative, you need to compute the final velocity with the information given.

pssh...I was thikning way too into it...thanks for your help.

I used the formula vf^2 = vi^2 + 2ad

Mathematically the answer turns out to be a positive number...Since it is still in the downward direction however, i should make it negative, correct?

EDIT: And thanks for moving the thread. Sry




What is the mathematical term for the path that the bullet takes? Projectile motion is always parabolic, however I am not able to see how that can be so with this problem...
 
Last edited:
If you were to graph it (distance vs time), youd see why it is parabolic. If the bullet was shot at 0 degrees (horizontally), youd see half the parabola. You would see half the parabolic curve because at this point the vertical velocity was 0 (which is also true when a projectile reaches the maximum height in its flight). In that case, its as if you are only seeing half the projectiles motion. A similar situation occurs in this example. However, the projectile does not start at the maximum height, it begins on its way back downwards. Thats why you would only see a certain portion of the parabola. In each situation, the projectile follows a parabolic curve. However, in the last 2 situations, it only doesn't do so for the entire curve. Does that make any sense?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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