How Fast Must a Puma Jump to Reach 11.5 Feet?

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SUMMARY

The discussion centers on calculating the initial velocity required for a puma to jump to a height of 11.5 feet (3.51 meters) at an angle of 43 degrees. The key equation used is V^2 = Vyo^2 + 2ay, where V is the final velocity (0 at maximum height), a is the acceleration due to gravity (-9.8 m/s²), and y is the change in height (3.51 m). Participants confirmed that the final velocity at the peak of the jump is zero, leading to the correct calculation of the initial velocity.

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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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