How Fast Must a Puma Jump to Reach 11.5 Feet?

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Homework Help Overview

The problem involves determining the initial speed required for a puma to jump to a height of 11.5 feet (3.51 meters) at an angle of 43 degrees. It is framed within the context of kinematics in two dimensions.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to use kinematic equations to find the initial velocity but expresses uncertainty about solving for time and initial velocity. Some participants suggest using a specific kinematic equation and confirm assumptions about final velocity at maximum height.

Discussion Status

Participants are exploring different kinematic equations and discussing the assumptions related to final velocity at the peak of the jump. Some guidance has been provided regarding the use of a specific equation, and there appears to be a productive exchange of ideas without a clear consensus on the approach.

Contextual Notes

The original poster notes the absence of horizontal distance in the problem, which adds to the complexity of finding a solution.

rcwha
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another 2 dimensional problem :(

The best leaper in the animal kingdom is the puma, which can jump to a height of 11.5ft (3.51m) when leaving the ground at an angle of 43 degress. With what speed, must the animal leave the ground to reach that height?

at first i thought i could find the time it took for the puma to get to it's maximum height by setting V=0 but i don't that's right...

i tried using the equation

change in y=(sin43)volt-.5g(t^2)

with change in y = 3.51 m


but i don't know how to solve for t or Vo

this question would be so much easier if the question gave the horizontal distance but it doesnt..so what do i do?
 
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You need a kinematic equation with inital velocity, final velocity, acceleration and displacement. Can you think of one?

HINT: You know the final velocity

~H
 
is it V^2=Vyo^2 + 2ay

a= -9.8
im assuming final velocity= 0 b/c that's what v is going to = at its max height

thank you...i ended up with the right answer
 
rcwha said:
is it V^2=Vyo^2 + 2ay

a= -9.8
im assuming final velocity= 0 b/c that's what v is going to = at its max height

thank you...i ended up with the right answer

Spot on. No problem :smile:

~H
 

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