To solve this problem we will use the kinematic equations of descriptive motion of a projectile for which both the height reached and the distance traveled are defined. From this type of movement the lion reaches a height (H) of 3m and travels a horizontal distance (R) of 10 m. Mathematically the equations that describe this movement are given as,
[tex]H = \frac{v_0^2sin^2\theta}{2g}[/tex]
[tex]R = \frac{v_0^2 sin 2\theta}{g}[/tex]
Dividing the two equation we have that
[tex]\frac{H}{R}=\frac{\frac{v_0^2sin^2\theta}{2g}}{\frac{v_0^2 sin 2\theta}{g}}[/tex]
[tex]\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{sin2\theta}[/tex]
[tex]\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{2sin\theta cos\theta}[/tex]
[tex]\frac{H}{R}= \frac{1}{4} \frac{sin\theta}{cos\theta}[/tex]
[tex]\frac{H}{R}= \frac{1}{4} tan\theta[/tex]
Substituting values of H and R, we get
[tex]\frac{3}{10} = \frac{1}{4} tan\theta[/tex]
[tex]\theta = tan^{-1} \frac{12}{10}[/tex]
[tex]\theta = 50.2\°[/tex]
Substituting the value of \theta in equation we get,
[tex]H = \frac{v_0^2sin^2\theta}{2g}[/tex]
[tex]v_0^2 = \frac{H 2g}{sin^2\theta}[/tex]
[tex]v_0^2 = \frac{3*2*9.8}{sin^2(50.2)}[/tex]
[tex]v_0^2 = 99.62[/tex]
[tex]v_0 = \sqrt{99.62}[/tex]
[tex]v_0 = 9.98m/s[/tex]
Therefore the speed of the mountain lion just as it leaves the ground is 9.98m/s at an angle of 50.2°
