The best leaper in the animal kingdom is the puma, which can jump to a height of 11.5 ft when leaving the ground at an angle of 52°. With what speed, in SI units, must the animal leave the ground to reach that height?

___m/s

Vi vertical= Speed*sin52

Vf^2=Vi^2+2gh where Vf (vertical velocity at the top) is zero.

100 m/s

To determine the speed at which the puma must leave the ground to reach a height of 11.5 ft (which is approximately 3.5052 meters), we can use the principles of projectile motion.

The key data we have is the height (h) and the launch angle (θ) of the puma. We can use the following formula to find the initial velocity (v):

h = (v^2 * sin^2(θ)) / (2 * g),

where:
- h is the height,
- v is the initial velocity,
- θ is the angle of launch, and
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

Rearranging the equation, we can solve for v:

v = √((h * 2 * g) / sin^2(θ)).

Now let's substitute the values given into the equation and calculate the speed:

h = 3.5052 meters,
θ = 52° (convert to radians: θ_rad = 52° * π / 180°),
g = 9.8 m/s^2.

v = √((3.5052 * 2 * 9.8) / sin^2(52°_rad)).

Calculating the expression in the square root:

v = √((68.62896) / sin^2(52°_rad)).

We need to find sin^2(52°_rad), which is:

sin^2(52°_rad) ≈ sin^2(0.9072) ≈ 0.821689.

Now substitute this value back into the equation:

v = √((68.62896) / 0.821689).

v ≈ √(83.6541).

Calculating the square root:

v ≈ 9.146 m/s.

Therefore, the puma must leave the ground with a speed of approximately 9.146 m/s to reach a height of 11.5 ft.