Can a Stuntman Jump Safely into a Pool from a 50-Meter Building?

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Homework Help Overview

The problem involves a stuntman attempting to jump from a 50-meter tall building into a pool located 21 meters horizontally from the base of the building. The discussion focuses on calculating the initial velocity required for the stuntman to land in the pool and evaluating the feasibility of such a jump.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for the final velocity using kinematic equations and question the choice of roots in the context of the stuntman's downward motion. There are also considerations about the realism of the required velocity for a human.

Discussion Status

The discussion is ongoing, with participants providing feedback on calculations and clarifying aspects of the mathematical approach. There is recognition of potential errors in reasoning, but no consensus has been reached regarding the final interpretation of the results.

Contextual Notes

Participants mention the impact of fatigue on their reasoning abilities, which may affect their calculations and understanding of the problem.

Alimarin
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Homework Statement


A stuntman must jump in a pool from a 50.0 meter tall building. The pool is a horizontal distance of 21.0 meters from the base of the building.

a) What is the initial velocity of the stuntman to be able to land in the pool?
b) Do you think this velocity is reasonable for a human?
c) Realistically, how tall must this building be for the stuntman to be able to make the landing safely?


Homework Equations


Vf^2 = Vi^2 + 2aΔx
Vf = Vi + aΔt
Xf-Xi = Δx
Given:
Xix = 0m
Xfx = 21m
Xiy = 50m
Xfy = 0m
Δx = 21m
Δy = -50m
a= -9.81m/s^2
Viγ = 0m/s

Relevant Unknowns:
Δt = ?
Vfγ = ?

The Attempt at a Solution


Vf^2 = Vi^2 + 2aΔx
Vf^2 = 0^2 + 2(-9.81m/s^2)(-50m)
Vf^2 = 2(9.81m/s^2)(50m)
Vf^2 = 981
Vf = 31.3 m/s

Vf = Vi + aΔt
31.3 = 0 + (-9.81m/s^2) Δt
31.3 = -9.81m/s^2 (Δt)
31.3 = Δt
-9.81m/s^2

Yeah, you see the issue, I made a stupid mistake and my brain is on 0 hours of sleep. and I have school in 1 hour.
 
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You were close, Vf^2 = 981 is right. But then you chose the positive root. The actual answer is the negative root Vf = -31.3 because the stuntman is falling downwards. So the change in time is the positive value of the answer you got.
 
BruceW said:
You were close, Vf^2 = 981 is right. But then you chose the positive root. The actual answer is the negative root Vf = -31.3 because the stuntman is falling downwards.

But mathematically that would take me into irrational numbers, E.g. 31.3i

Which would mean I would have to do some algebraic things I don't even remember how to do anymore.

I just realized that square roots have a + or -, never mind.

See? My brain doesn't have it's sensitivity to serotonin at the moment.
 
No worries, good luck finishing the question.
 
Alright, this time I did it right.

Garcias
 

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