Mayan kings and many school sports teams are named for the puma, cougar, or mountain lion Felis concolor, the best jumper among animals. It can jump to a height of 12.2 ft when leaving the ground at an angle of 44.2°. With what speed, in SI units, does it leave the ground to make this leap?

2*9.8*12.2 = (vo sin44)^2

Have to convert the 12.2 to meters

To determine the speed at which the puma leaves the ground to make the leap, we can use the principles of projectile motion. Projectile motion involves the motion of an object through the air under the influence of gravity.

In this case, we know the vertical height reached by the puma is 12.2 ft, and the launch angle is 44.2°. We need to convert these measurements to SI units for consistency. 1 ft is approximately equal to 0.3048 meters.

So, the vertical height reached by the puma is (12.2 ft) * (0.3048 m/ft) = 3.71856 meters.

Now, using the principles of projectile motion, we can calculate the initial vertical velocity (v_y) required for the puma to reach that height.

The equation we can use is:

v_y^2 = v_0^2 * sin^2(theta) - 2 * g * h

where:
v_y is the vertical component of the initial velocity,
v_0 is the initial velocity (which we need to find),
theta is the launch angle (44.2°),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
h is the vertical height reached (3.71856 meters).

Rearranging the equation, we can solve for v_0:

v_0 = sqrt((v_y^2 + 2 * g * h) / sin^2(theta))

Since the puma leaves the ground, the vertical component of the initial velocity (v_y) is 0 m/s.

Now, let's plug in the values and calculate the speed:

v_0 = sqrt((0 + 2 * 9.8 * 3.71856) / sin^2(44.2°))
= sqrt(72.9016 / (sin(44.2°))^2)

Using a calculator, we find:

v_0 ≈ 15.16 m/s

Therefore, the puma leaves the ground with a speed of approximately 15.16 m/s to make the leap.

To find the speed at which the puma leaves the ground to make the leap, we can use the laws of projectile motion.

The vertical component of the velocity can be determined using the equation:

Vf = Vi + at

Where:
Vf is the final vertical velocity (which will be zero when the puma reaches its maximum height),
Vi is the initial vertical velocity,
a is the acceleration (in this case, acceleration due to gravity, which is approximately -9.8 m/s²), and
t is the time it takes to reach the maximum height (which we can find using the equation Vf = Vi + at, but substituting Vf = 0).

Let's find the initial vertical velocity (Vi) using the given information:

Vi = V * sin(θ)

Where:
V is the total velocity, and
θ is the launch angle (44.2° in this case).

Now, let's find the total velocity (V) using the given information:

V = d / t

Where:
d is the vertical distance covered (12.2 ft, which is approximately 3.72 m), and
t is the time it takes to reach the maximum height.

Finally, we can calculate the time it takes to reach the maximum height using the equation:

Vf = Vi + at

Since Vf = 0 (at the maximum height) and a = -9.8 m/s², we can find t.

Let's proceed with the calculations:

Step 1: Find the total velocity (V)
V = d / t
V = 3.72 m / t

Step 2: Find the time to reach the maximum height (t) using Vf = Vi + at
0 = Vi + (-9.8 m/s²) * t

Step 3: Find the initial vertical velocity (Vi) using Vi = V * sin(θ)
Vi = V * sin(44.2°)

Step 4: Substitute the value of Vi from Step 3 into the equation from Step 2:
0 = V * sin(44.2°) - 9.8 m/s² * t

Now, we can solve this system of equations to find the value of t and subsequently the speed at which the puma leaves the ground:

Step 5: Solving for t in the equation from Step 2:
0 = V * sin(44.2°) - 9.8 m/s² * t
t = V * sin(44.2°) / 9.8 m/s²

Step 6: Substitute the expression for t from Step 5 into the equation from Step 1:
V = 3.72 m / (V * sin(44.2°) / 9.8 m/s²)
V = 3.72 m * 9.8 m/s² / (V * sin(44.2°))
V² = (3.72 m * 9.8 m/s²) / sin(44.2°)
V = √((3.72 m * 9.8 m/s²) / sin(44.2°))

Using a calculator, the speed at which the puma leaves the ground can be calculated using the given data.