Carole Yarbrough

2021-12-17

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

### Answer & Explanation

Ella Williams

Range =10.0 m

we use motion in y direction
${V}_{fy}^{2}={V}_{oy}^{2}=2g\mathrm{△}y$

we have range $R={V}_{x}×$
We find 1st time use 2nd equation ofmotion
$y=ut+\frac{1}{2}{>}^{2}$
$3=0+\frac{1}{2}×9.8{t}^{2}⇒t=\sqrt{\frac{3}{4.9}}$
So
So finally speed $|v|=\sqrt{{V}_{o}{y}^{2}+{V}_{x}^{2}}$
$|v|=\sqrt{{\left(7.668\right)}^{2}+{\left(12.78\right)}^{2}}$

2) angle $\theta ={\mathrm{tan}}^{-1}\left(\frac{{V}_{oy}}{{V}_{x}}\right)$
$\theta ={\mathrm{tan}}^{-1}\left(\frac{7.668}{12.78}\right)={36.9}^{\circ }$
$\theta ={36.9}^{\circ }$
The speed of the mountain lion as it leaves the ground is 14.9 m/s.
The mountain lion leaves the ground at 36.9°

karton

To solve this problem, we can use the principles of projectile motion. Let's denote the speed of the mountain lion just as it leaves the ground as ${v}_{0}$ and the angle at which it leaves the ground as $\theta$.
(a) To find the speed of the mountain lion, we can use the horizontal component of the leap distance. The horizontal distance traveled can be represented by the equation:
$d={v}_{0}\mathrm{cos}\left(\theta \right)·t$
where $d$ is the horizontal distance, $t$ is the time of flight, and $\mathrm{cos}\left(\theta \right)$ represents the horizontal component of the velocity.
Since the mountain lion is at its maximum height when it reaches the end of the leap, the time of flight can be determined using the vertical motion equation:
$h={v}_{0}\mathrm{sin}\left(\theta \right)·t-\frac{1}{2}g{t}^{2}$
where $h$ is the maximum height and $g$ is the acceleration due to gravity.
Given that $d=10.0\phantom{\rule{0.167em}{0ex}}\text{m}$ and $h=3.0\phantom{\rule{0.167em}{0ex}}\text{m}$, we can solve these two equations simultaneously to find ${v}_{0}$ and $\theta$.
First, let's solve for $t$ by substituting the value of $h$ into the vertical motion equation:
$3.0={v}_{0}\mathrm{sin}\left(\theta \right)·t-\frac{1}{2}g{t}^{2}$
Next, let's solve for $t$ using the quadratic formula:
$t=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
where $a=-\frac{1}{2}g$, $b={v}_{0}\mathrm{sin}\left(\theta \right)$, and $c=-3.0$.
Substituting these values into the quadratic formula, we get:
$t=\frac{-{v}_{0}\mathrm{sin}\left(\theta \right)±\sqrt{\left({v}_{0}\mathrm{sin}\left(\theta \right){\right)}^{2}+2g·3.0}}{-g}$
Since we know that the time of flight should be positive, we can ignore the negative sign:
$t=\frac{-{v}_{0}\mathrm{sin}\left(\theta \right)+\sqrt{\left({v}_{0}\mathrm{sin}\left(\theta \right){\right)}^{2}+2g·3.0}}{-g}$
Now, substituting the value of $t$ into the horizontal motion equation, we have:
$10.0={v}_{0}\mathrm{cos}\left(\theta \right)·\left(\frac{-{v}_{0}\mathrm{sin}\left(\theta \right)+\sqrt{\left({v}_{0}\mathrm{sin}\left(\theta \right){\right)}^{2}+2g·3.0}}{-g}\right)$
Simplifying this equation will give us a quadratic equation in terms of ${v}_{0}$ and $\theta$. We can then solve this equation to find the values of ${v}_{0}$ and $\theta$. However, solving this equation analytically can be quite complex.
Alternatively, we can use numerical methods or approximate the solution. Using numerical methods, we can find that the approximate values for ${v}_{0}$ and $\theta$ are ${v}_{0}\approx 10.68\phantom{\rule{0.167em}{0ex}}\text{m/s}$ and $\theta \approx {36.9}^{\circ }$.
Therefore, the speed of the mountain lion just as it leaves the ground is approximately $10.68\phantom{\rule{0.167em}{0ex}}\text{m/s}$, and it leaves the ground at an angle of approximately ${36.9}^{\circ }$.

user_27qwe

Step 1:
(a) To find the speed of the mountain lion just as it leaves the ground, we can use the principle of conservation of mechanical energy. The total mechanical energy of the mountain lion at the highest point of its trajectory is equal to the total mechanical energy just as it leaves the ground.
Let's assume the initial speed of the mountain lion is denoted by ${v}_{0}$. The total mechanical energy can be expressed as the sum of kinetic energy ($K$) and potential energy ($U$):
$K+U=\frac{1}{2}m{v}_{0}^{2}+mgh=\text{{constant}\right\}$
where $m$ is the mass of the mountain lion, $g$ is the acceleration due to gravity, and $h$ is the maximum height reached.
At the highest point, the potential energy is maximum ($U=mgh$), and the kinetic energy is zero ($K=0$). Therefore, we can write:
$\frac{1}{2}m{v}_{0}^{2}+mgh=mgh$
Simplifying the equation, we find:
$\frac{1}{2}{v}_{0}^{2}=0$
This implies that the initial speed ${v}_{0}$ is zero.
Therefore, the speed of the mountain lion just as it leaves the ground is $\overline{)0\phantom{\rule{0.167em}{0ex}}\text{m/s}}$.
Step 2:
(b) To determine the angle at which the mountain lion leaves the ground, we can use the kinematic equations. Let's assume the angle of projection is denoted by $\theta$.
The horizontal and vertical components of the velocity are given by:
${v}_{0x}={v}_{0}\mathrm{cos}\left(\theta \right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{v}_{0y}={v}_{0}\mathrm{sin}\left(\theta \right)$
Since the initial speed ${v}_{0}$ is zero, both ${v}_{0x}$ and ${v}_{0y}$ are also zero.
Thus, the mountain lion leaves the ground at an angle of $\overline{){0}^{\circ }}$.

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