What is the maximum height reached by a launched human?

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically calculating the maximum height reached by a human launched vertically at a certain speed. The original poster presents a scenario with a 70kg human being launched at a 90-degree angle with an initial velocity of 15 mph.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations to determine the height but seems to struggle with the correct application of acceleration and velocity. Some participants question the assumptions made about the initial conditions and the forces acting on the human after launch.

Discussion Status

Participants are actively engaging in clarifying the physics concepts involved, with some suggesting that the original poster reconsider their approach to the equations used. There is a recognition of the need to properly account for gravitational acceleration and its effect on the motion of the launched human.

Contextual Notes

There is a discussion about unit conversions between metric and imperial systems, as well as the need to correctly apply the principles of projectile motion without introducing incorrect multiplications or assumptions about forces.

Richard09876
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Homework Statement



Hi guys, I am working through a very long problem and I am stuck at this part.

So I have a 70kg human being launched at a 90 degree angle at 15mph. I need to figure out how high in the air he will get before velocity reaches zero again and he begins to fall.

Homework Equations


Vf = Vi + at

Rearranged to: a = (Vf-Vi)/ t

A- time
Vi- initial velocity
Vf- final velocity
t- -9.8m/s^2

The Attempt at a Solution



After I rearranged the problem I got a = (0-2.5)/-9.8

a = .255

Now I used D= Vi*t (1/2) a t ^2

6.25 (.255) = 1.59

1.59 + 1/2 (-9.81)(.255)= 1.27 mIs this correct??[/B]
 
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Not even close.

Since the human is being launched into the air (presumably without a rocket attached to his back side), he is not accelerating off the ground, but has a constant velocity of 15 mph as he leaves the ground. The only thing retarding his upward movement is the gravitational pull of earth.

What is g for Earth in foot-second units?

BTW: I've moved this thread to the Intro Physics HW forum, which is more appropriate given the nature of the problem.
 
It would be 9.8m/s^2

So, -32.15f/s^2 right?

Now I convert 15mph to 22 f/s.

Now I multiply 22f/s x 32.1 f/s^2 = 706 f/s^2

Is this right?
 
Richard09876 said:
It would be 9.8m/s^2

So, -32.15f/s^2 right?

Now I convert 15mph to 22 f/s.

Now I multiply 22f/s x 32.1 f/s^2 = 706 f/s^2

Is this right?

You're going from bad to worse.

Why are you multiplying the acceleration due to gravity by the initial velocity of the human?

For the record, the units of ft / s multiplied by ft / s2 are not equal to ft / s2.
 
I am really trying hard to work through this one. Could you suggest a direction I should take?

Perhaps I should not be multiplying at all. Since the acceleration due to gravity is 9.8ms^2 I should subtract that from the 22f/s since the gravity is "retarding" the human. Am I getting anywhere here?
 
Richard09876 said:
I am really trying hard to work through this one. Could you suggest a direction I should take?

Perhaps I should not be multiplying at all. Since the acceleration due to gravity is 9.8ms^2 I should subtract that from the 22f/s since the gravity is "retarding" the human. Am I getting anywhere here?

You're not getting anywhere because you aren't using the correct equation to solve this problem. You want to select the kinematic equation or equations which will allow you to calculate the distance traveled by the human after he is launched.

http://www.ronknott.com/MEI/MechSuvatEquns.html
 

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