Odd projectile motion question.

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Homework Help Overview

The discussion revolves around a projectile motion problem involving the calculation of maximum height based on a given time of flight of 3.4 seconds and gravitational acceleration of 9.8 m/s².

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of kinematic equations, questioning the need for specific formulas and the interpretation of time at the peak of the trajectory. Some suggest halving the time of flight to find the time to maximum height, while others discuss the implications of free fall after reaching that height.

Discussion Status

The discussion is active, with participants offering various approaches to the problem. Some have attempted calculations based on their interpretations of the motion, while others are clarifying concepts related to the trajectory and vertical motion. There is no explicit consensus on the correctness of the calculations presented.

Contextual Notes

Participants are working within the constraints of the problem statement and are questioning the assumptions made regarding the motion of the projectile and the application of kinematic equations.

letsfailsafe
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Homework Statement


Question: "The time of flight of the lure through the air is 3.4 seconds. (g= 9.8m/s/s)
Calculate the maximum height of the lure in its projectile motion."


Homework Equations


Not sure if I need to use this:
d = (u)(t) + (1/2)(a)(t^2)


The Attempt at a Solution



I don't know how this really works...

I'm guess to half the time (3.4/2). Thats the time when the object is at rest for a very short time (initial velocity would be 0). And use that equation on to solve the d.
 
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Once the object is at rest (top of the trajectory) what follows (on the vertical axis) is just a free fall, isn't it?
 
You could start by finding the initial vertical component of velocity and using another kinematic relation to find the height. Or do as Borek suggested.
 
So...

d = (0)(1.6) + (1/2)(9.8)(1.6)^2

= 12.544 m

Is it right?
 
The time of flight of the lure through the air is 3.4 seconds means the lure left the Earth at t=0 and landed back to Earth at 3.4sec.

So vertically the final position is back to where it started.
Y position is a horizontal line parallel to x-axis.
 
letsfailsafe said:
So...

d = (0)(1.6) + (1/2)(9.8)(1.6)^2

= 12.544 m

Is it right?
Right idea, but 3.4/2 ≠ 1.6. :wink:
 

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