How fast must a Puma leave to ground to go 12 feet high?

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To determine the speed a Puma must leave the ground to reach a height of 12 feet at a 45-degree angle, the discussion emphasizes the need to apply kinematic formulas for projectile motion. The trajectory can be analyzed using the equation for parabolic motion, considering both vertical and horizontal components separately. At the peak height, the vertical component of velocity will be zero, which is a crucial aspect of the calculation. The conversation highlights the importance of understanding the relationship between angle and velocity in projectile motion. Overall, applying these principles will lead to the correct solution for the Puma's launch speed.
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This seems like an easy problem but without "time" i"m confused.

The best leaper is a Puma that can jump to a height of 12 feet when leaving the ground at a 45 degree angle. With what speed in SI units must the Puma leave the ground to reach this height?

I'm not sure what formula to apply since there isn't any time mentioned.
 
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This is interesting.. because cos(45^o) = sin(45^o) :smile:
 
Do you know the formula's for the two dimensional movement under the gravity-force. You have a formule that describes the trajectory (a parabole) y = f(x). Now since the angle is 45° you can calculate all you need using this formula. There is another option when you use the x and y components for the position and velocity. Keep in mind that the y-component of the velocity is zero at the point of maximal height...

bonne chance

regards
marlon
 
Correct. Almost always in projectiles you must consider the vertical and horizontal components of the trajectory of an object separately. You can solve this problem using marlon's hints, kinematics formulas, and physical thinking. :smile:
 
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