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7 days ago

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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?

1 answer:

ValentinkaMS [1.1K]7 days ago

5 0

To tackle this question we will apply the kinematic equations that describe the motion of a projectile, where both maximum height and distance traveled are defined. This scenario demonstrates a lion achieving a height (H) of 3m and covering a horizontal distance (R) of 10m. The equations governing this kind of motion are expressed as follows:

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

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

By dividing the two equations, we determine:

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

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

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

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

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

Plugging in the values for H and R yields:

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

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

$\theta = 50.2\°$

After substituting \theta into the relevant equation, we find:

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

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

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

$v_0^2 = 99.62$

$v_0 = \sqrt{99.62}$

$v_0 = 9.98m/s$

In conclusion, the mountain lion's launch speed upon takeoff is approximately 9.98m/s at an angle of 50.2°.

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Refer to the attached diagram for enhanced clarity regarding the question.

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