Projectile Motion Time Question

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Homework Help Overview

The discussion revolves around projectile motion, specifically the time it takes for an object to ascend to its peak and descend back to the same height. The original poster questions the symmetry of time in projectile motion, noting a scenario where an object launched by a slingshot reaches its peak in about 1 second but takes approximately 2 seconds to return to the ground, suggesting the influence of factors like air resistance and wind.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the possibility of differing ascent and descent times due to real-world factors such as air resistance and changes in drag coefficient. Questions arise regarding how terminal velocity affects the motion, particularly in the vertical direction.

Discussion Status

Some participants have provided insights into potential causes for the observed time discrepancy, including the effects of drag and terminal velocity. There is ongoing inquiry into the implications of these factors on the motion of the projectile, with no explicit consensus reached yet.

Contextual Notes

Participants are considering the effects of air resistance and terminal velocity in a real-world context, which may not align with idealized projectile motion assumptions. The original poster's scenario suggests a need for clarification on these influences.

Nerdyboy
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Homework Statement


In projectile motion, the time it takes for an object to get from a certain height to the peak is the same time as the time to reach the same height from the peak, correct? Well then, I came across something in which an object was shot in the air via a slingshot and peaked at about 1 second. Then, it took approximately 2 seconds to return to the ground at the same height. Is this possible? If so, how? (Lets say that there may be air resistance and wind as an influence)


Homework Equations


Kinematics Equations:
V = Vo +at
x = 1/2(Vo+V)t
x = Vit + 1/2at^2
V^2 = Vo^2 + 2ax

The Attempt at a Solution


I thought that when in projectile motion, an object would have the same time reaching the peak as it would reaching the ground from the peak. (Symmetry).
 
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Nerdyboy said:
Well then, I came across something in which an object was shot in the air via a slingshot and peaked at about 1 second. Then, it took approximately 2 seconds to return to the ground at the same height. Is this possible? If so, how? (Lets say that there may be air resistance and wind as an influence)
I'm guessing your talking about real world measurements. Let's see, this might be caused by a change in the drag coefficient (eg. as an extreme example, shooing a parachute in the air while balled up, then letting it unfurl as it fell), or a wind gust. It also might happen if the object was shot faster than terminal velocity, since the speed of the object wouldn't exceed terminal velocity on its return trip to the ground.
 
Sciurus said:
I'm guessing your talking about real world measurements. Let's see, this might be caused by a change in the drag coefficient (eg. as an extreme example, shooing a parachute in the air while balled up, then letting it unfurl as it fell), or a wind gust. It also might happen if the object was shot faster than terminal velocity, since the speed of the object wouldn't exceed terminal velocity on its return trip to the ground.
One more question then, how does terminal velocity affect the object in projectile motion. Does it slow down the velocity in the y direction of the projectile because I know I am almost positive it won't affect the x velocity.
 
can someone help me with the post need help...please...
 
Nerdyboy said:
One more question then, how does terminal velocity affect the object in projectile motion. Does it slow down the velocity in the y direction of the projectile because I know I am almost positive it won't affect the x velocity.
Terminal velocity is the speed a falling object tends toward -- it's the speed where the downward force of gravity is equal to the upward force due to drag.

Drag -- much like friction -- slows an object down, so if the object has an x and y component of velocity, it would reduce both. In a similar way, if you drew x and y-axis on a table top, then slid a note card across table, the notecard's x and y velocity would slow down with friction.

But the drag force also depends on the speed (both x and y), so it's a bit harder to work with in two dimensions.

The get an idea of how it works, consider shooting an object straight up, so we only have to work on one dimension. We'll first consider the fall from the maximum height. The object accelerates downward because of gravity, but as it starts moving faster, the drag increases, providing an upward force. The end result is that the object approaches its terminal velocity, and gets very close, but doesn't go any faster. Not since you can shoot the object faster than the terminal velocity, you can get it up to the highest point faster than it comes down.
 

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