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 11.2 ft when leaving the ground at an angle of 43.8°. With what speed, in SI units, does it leave the ground to make this leap?
To calculate the speed at which the puma leaves the ground, we can use the principles of projectile motion. The initial vertical velocity (Vy) can be determined using the equation:
Vy = V0 * sin(θ)
Where:
- Vy is the vertical component of velocity
- V0 is the initial velocity (speed)
- θ is the launch angle
We are given:
- Height (H) = 11.2 ft = 3.413 m
- Launch angle (θ) = 43.8°
First, we need to convert the height to meters to be consistent with SI units. We know that 1 ft = 0.3048 m, so:
H = 11.2 ft * 0.3048 m/ft ≈ 3.413 m
Now, we can solve for Vy:
Vy = V0 * sin(θ)
Rearranging the equation to solve for V0 gives us:
V0 = Vy / sin(θ)
Substituting the known values:
V0 = 3.413 m / sin(43.8°)
Using a scientific calculator, we can evaluate sin(43.8°) to be approximately 0.6820, so:
V0 ≈ 3.413 m / 0.6820 ≈ 5.0 m/s
Therefore, the cougar leaves the ground with a speed of approximately 5.0 m/s when making this leap.