How Far Does a 500 m/s Projectile Travel Horizontally at -50m Elevation?

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Homework Help Overview

The problem involves a projectile launched at an initial velocity of 500 m/s at a 30º angle above the horizontal. The task is to determine the horizontal distance traveled when the projectile is 50 m below the launch point, as well as the maximum height achieved and the angle of the velocity vector at that point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss breaking down the initial velocity into horizontal and vertical components to analyze the projectile's motion. There is mention of calculating the time to reach -50 m elevation and how that relates to horizontal distance traveled. Some participants express uncertainty about the effects of gravity on horizontal motion and question assumptions about vector independence.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of projectile motion and vector resolution. Some guidance has been offered regarding the independence of horizontal and vertical motions, but there is no explicit consensus on the approach to take or the calculations needed.

Contextual Notes

Participants note the complexity of the problem due to the initial conditions and the need to consider the effects of gravity on vertical motion while maintaining horizontal speed. There is an acknowledgment of potential confusion regarding the influence of gravity on horizontal velocity.

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Homework Statement



Determine how far away the projectile is from the launch point when it is 50m below the launch point. Determine the max height achieved by the projectile. What is the angle that the velocity vector makes with the horizontal at the point where the projectile is -50m from launch point. V=500 m/s @ 30º angle. What is distance in horizontal and angle of velocity vector when at -50m from launch.

Homework Equations



v = u + at - Vf^2 - Vo^2 = 2aD

The Attempt at a Solution



Having trouble starting this problem...
 
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I am having a little trouble following what you wrote. If I am interpreting correctly. This is a projectile launched from an elevation with an initial velocity of 500 m/s, directed 30º above the horizontal. You are asked to calculate information about the projectile when it has dropped 50 m below the launch point (and gone some distance forward at the same time).

You can approach most two-dimensional motion problems by resolving vectors into perpendicular components. In this case that means taking the 500 m/s directed up at 30º and breaking it into a vertical component and a horizontal component. You can then solve for how long it takes the projectile to rise to its peak and fall to -50 m. That is the same time the projectile moves horizontally at a constant speed (since there are no forces acting on it in the horizontal direction).
 
well, you need to take into account the "speed" of gravity and how much it will slow the velocity down every second.

Putting the into in a graph will help, or even better... put the info in a graphics calculator and it will calculate the max height ONLY!

for the horizontal direction, I'm not too sure?

@Fewmet- wouldn't gravity affect the horizontal speed because of the angles of the vectors?
 
miniradman said:
@Fewmet- wouldn't gravity affect the horizontal speed because of the angles of the vectors?

No. I know that it seems, intuitively, like the downward pull of gravity makes something like a horizontally thrown ball speed up. What is increasing it the sum of the horizontal and vertical velocities.

Consider tossing a coin vertically while on a uniformly moving train. If gravity caused a horizontal acceleration, the coin would move forward faster than the train, than you and than your hand. It would land in front you on the train. The fact that it falls back into your hand is consistent with the principle that perpendicular vectors act independently of each other.
 

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