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

1. Oct 1, 2004

### Physically Impaired

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.

2. Oct 1, 2004

### Pyrrhus

This is interesting.. because $$cos(45^o) = sin(45^o)$$

3. Oct 1, 2004

### marlon

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

4. Oct 1, 2004

### Sirus

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.