Question

A cougar jumps to a height of 3.15 m when leaving the ground at an angle of 42.8°. what is its initial speed (in m/s) as it leaves the ground?

Answer 1

The initial speed of the cougar as it leaves the ground is approximately 6.24 m/s.

Step-by-step explanation:

To find the initial velocity of the cougar, we can use the energy conservation principle. The initial potential energy when the cougar jumps is given by mgh, where m is the mass of the cougar, g is the acceleration due to gravity, and h is the vertical height. The initial kinetic energy is given by (1/2)mv^2, where v is the initial velocity. Since energy is conserved, we can equate the potential energy to the kinetic energy and solve for v. In this case, the equation is mgh = (1/2)mv^2. We can cancel out the mass and solve for v by taking the square root of (2gh). Substituting the given values for g and h, we have v = sqrt(2 * 9.8 * 3.15) ≈ 6.24 m/s.