Question

A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion just as it leaves the ground? At what angle does it leave the ground?

Answer 1

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,

$H = \frac{v_0^2sin^2\theta}{2g}$

$R = \frac{v_0^2 sin 2\theta}{g}$

Dividing the two equation we have that

$\frac{H}{R}=\frac{\frac{v_0^2sin^2\theta}{2g}}{\frac{v_0^2 sin 2\theta}{g}}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{sin2\theta}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{2sin\theta cos\theta}$

$\frac{H}{R}= \frac{1}{4} \frac{sin\theta}{cos\theta}$

$\frac{H}{R}= \frac{1}{4} tan\theta$

Substituting values of H and R, we get

$\frac{3}{10} = \frac{1}{4} tan\theta$

$\theta = tan^{-1} \frac{12}{10}$

$\theta = 50.2\°$

Substituting the value of \theta in equation we get,

$H = \frac{v_0^2sin^2\theta}{2g}$

$v_0^2 = \frac{H 2g}{sin^2\theta}$

$v_0^2 = \frac{3*2*9.8}{sin^2(50.2)}$

$v_0^2 = 99.62$

$v_0 = \sqrt{99.62}$

$v_0 = 9.98m/s$

Therefore the speed of the mountain lion just as it leaves the ground is 9.98m/s at an angle of 50.2°