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Korvikt [17]
3 years ago
10

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 j

ust as it leaves the ground? At what angle does it leave the ground?
Physics
1 answer:
nika2105 [10]3 years ago
5 0

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°

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