How Fast Must a Puma Jump to Reach 11.5 Feet?

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A puma can jump to a height of 11.5 feet (3.51 meters) at an angle of 43 degrees. To determine the initial speed required for this jump, the equation used is V^2 = Vyo^2 + 2ay, where the acceleration due to gravity is -9.8 m/s² and the final velocity at maximum height is 0. The discussion highlights the challenge of solving for time and initial velocity without horizontal distance information. Ultimately, the correct application of kinematic equations leads to the right answer for the initial velocity needed for the puma's jump.
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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)Vot-.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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