A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.

When the lion is at the maximum height, its vertical velocity is 0 m/s. Let’s use the following equation to determine the initial vertical velocity.

vf^2 = vi^2 + 2 * g * d

0 = vi^2 + 2 * -9.8 * 3

vi = √58.8

This is approximately 7.7 m/s. This happens during one half of the total time. Let’s use the following equation to determine this time.

vf = vi – g * t

0 = √58.8 – 9.8 * t

t = √58.8 ÷ 9.8

This is approximately 0.78 seconds. The total time is twice this time. During the total time, the lion moves a horizontal distance of 10 meters. Let’s use the following equation to determine the horizontal component of the lion’s initial velocity.

d = v * t

10 = v * 2 * √58.8÷ 9.8

v = 49 ÷ √58.8

This is approximately 6.4 m/s.

Speed = √[58.8 + (49 ÷ √58.8)^2]

This is approximately 10 m/s. To determine the angle from horizontal, use the following equation.

Tan θ = Vertical ÷ Horizontal

Tan θ = √58.8 ÷ (49 ÷ √58.8) = 1.2

The answer is 50°.

Ibelieve the correct answer is c