The best leaper in the animal kingdom is the puma, which can jump to a height of 12ft. when leaving the ground

Question

The best leaper in the animal kingdom is the puma, which can jump to a height of 12ft. when leaving the ground at an angle of 45 degrees. With what speed, in SI units, must the animal leave the ground to reach that height?

Answer 1

Ok
45 degress will also result to maximum distance as well So the pume does not follow maximm efficiency rule.

in any case.
as it jumps it has 2 velocities one in the vertical *y axis) and one in the horixontal X axis.

the Y axis under interest velocity is from treigonomeetric functions

Ux=sin45 Uo. threfore Uo=2Ux.

NOw 12ft = 3.6576m
so sim,ply you need the velocity that it will make an object to move for 12 before reaching a stop under a deceleration due to g.

now due to low of energy conservation the potential energy on the top (rest) must equal to the kinetic energy in the ground therefore :

1/2 mu^2 = mgh => u=SRT(2gh) where h= height in meters !

now this U is the Ux
you know that 2Ux = Uo.


so........

Uo is known.

Answer 2

Emmanuel P made a mistake in his trigonometry.
He said "Ux=sin45 Uo. threfore Uo=2Ux." That's wrong.
Sin45 is .707 so Uo=1.414Ux.

The puma has velocity V at a 45 degree angle. The vertical component is
Vy = .707*V
I agree that 12ft = 3.6576m.

And I agree with using the conservation of energy method.
(1/2)*m*Vy^2 = mgh
Vy^2 2*g*h so Vy = SRT(2*g*h) where h = 3.6576m.

This gives you Vy,
V = Vy / .707