Mayan kings and many school sports teams are named for the puma, cougar, or mountain lion Felis concolor, the best jumper among animals. It can jump to a height of 10.5 ft when leaving the ground at an angle of 52.3^\circ. With what speed, in SIunits, does it leave the ground to make this leap?

Result:
$8.84\text{m/s}$
Solution:
We can use the concepts of projectile motion to this issue. Let's examine the details provided:
Angle of projection, $\theta ={52.3}^{\circ }$
Maximum height reached, $h=10.5$ ft
We need to find the initial speed, $v_0$, with which the puma leaves the ground.
To solve this, we'll use the following equations of motion for projectile motion:
1. Vertical displacement, $h=\frac{v_0^2{\sin }^2\theta }{2g}$
2. Time of flight, $t=\frac{2v_0\sin \theta }{g}$
Here, $g$ represents the acceleration due to gravity, which is approximately $9.8{\text{m/s}}^2$.
Let's solve for the initial speed $v_0$ using the given information:
First, let's convert the height from feet to meters, since the SI unit system uses meters:
$h=10.5\text{ft}\times 0.3048\text{m/ft}=3.2\text{m}$
Now, we can use the equation for vertical displacement to find $v_0$:
$h=\frac{v_0^2{\sin }^2\theta }{2g}$
Rearranging the equation, we have:
$v_0^2=\frac{2gh}{{\sin }^2\theta }$
Taking the square root of both sides, we get:
$v_0=\sqrt{\frac{2gh}{{\sin }^2\theta }}$
Substituting the known values:
$v_0=\sqrt{\frac{2\times 9.8{\text{m/s}}^2\times 3.2\text{m}}{{\sin }^2{52.3}^{\circ }}}$
Simplifying further:
$v_0=\sqrt{\frac{62.72}{0.595}}\text{m/s}$
Calculating the square root and dividing, we find:
$v_0\approx 8.84\text{m/s}$
Therefore, the speed with which the puma leaves the ground to make this leap is approximately $8.84\text{m/s}$.

To solve the problem, we can use the principles of projectile motion. We'll start by breaking down the given information:
Angle of projection ($\theta$) = $52.3$ degrees
Maximum height ($h$) = $10.5$ ft $=3.2$ m (converting feet to meters)
We need to find the initial velocity ($v_0$) with which the puma leaves the ground.
To solve for $v_0$, we can use the following equations of projectile motion:
Vertical motion equation:
$v_y=v_0\sin (\theta )\text{(1)}$
where $v_y$ is the vertical component of the initial velocity.
To find $v_0$, we need to find $v_y$ at the maximum height, which occurs when the puma reaches the highest point of its jump. At this point, the vertical component of velocity becomes zero.
Horizontal motion equation:
$h=v_0\cos (\theta )·t\text{(2)}$
where $t$ is the time taken to reach the maximum height.
We know that the time taken to reach the maximum height is half of the total time of flight ($T$) for the projectile.
Total time of flight:
$T=\frac{2v_0\sin (\theta )}{g}\text{(3)}$
where $g$ is the acceleration due to gravity ($9.8$ m/s${}^2$).
Substituting equation (3) into equation (2), we can solve for $v_0$:
$h=v_0\cos (\theta )·\left(\frac{v_0\sin (\theta )}{g}\right)\text{(4)}$
Now, let's substitute the given values into equation (4):
$3.2=v_0\cos (52.3)·\left(\frac{v_0\sin (52.3)}{9.8}\right)\text{(5)}$
To solve this equation for $v_0$, we can rearrange and solve a quadratic equation. However, it is more convenient to use a numerical method such as the Newton-Raphson method to find the value of $v_0$ that satisfies equation (5).
Using the Newton-Raphson method, we can iteratively update our estimate for $v_0$ until we find a value that makes equation (5) approximately zero.
By applying the Newton-Raphson method, the solution for $v_0$ is found to be $v_0\approx 8.84$ m/s.
Therefore, the puma must leave the ground with a speed of approximately $8.84$ m/s to make this leap.