Question

Mayan kings and many school sports teams are named for the puma, cougar, or mountain lionFelis concolorthe best jumper among animals. It can jump to a height of$\mathbf{12}.\mathbf{0}\mathbf{ft}$when leaving the ground at an angle of$\mathbf{45}.\mathbf{0}°$. With what speed, inunits, does it leave the ground to make this leap?

Short answer

The initial velocity of the speed is $v_i=12\text{m}/\text{s}$.

Step-by-step solution

Definition of initial velocity.

• The initial velocity, denoted by $u$, is the velocity at time interval $t=0$. It's the speed at which the motion begins.
• The starting velocity is articulated as $u=v-at$if time, acceleration, and final velocity are provided.
• Where $u$is the initial velocity , $v$is the final velocity,is the time taken, and $a$is the acceleration

Find the time required to reach this height.

Given data,

$$\begin{gathered} y_{\max }=12\text{ft} \\ {y}_{\max }=\frac{12}{3.28} \\ {y}_{\max }=3.66\text{m} \end{gathered}$$

First, we need to find the time required to reach this height which we get by using this equation:

$v_{yf}=v_{yi}+a_{y}t$

Solving $t$for taking into account that at the maximum height $v_y=0$and $a=-g:$

$$\begin{gathered} t=\frac{v_{yf}-v_{yi}}{a_y} \\ t=\frac{0-v_{yi}}{-g} \\ t=\frac{v_{yi}}{g} \end{gathered}$$

Find the vertical displacement.

Then, we need to find the vertical displacement made in this time by using the following equation:

$$\begin{gathered} y_{\max }=v_{y,avg}t \\ {y}_{\max }=\left(\frac{v_{yi}+v_{yf}}{2}\right)t \\ {y}_{\max }=\left(\frac{v_{yi}}{2}\right)\left(\frac{v_{yi}}{g}\right) \\ {y}_{\max }=\frac{v_{yi}^2}{2g} \end{gathered}$$

Determine the initial velocity.

By knowing the angle of projection , we can find the initial velocity by using the following relation:

Therefore, the initial velocity of the speed is $v_i=12\text{m}/\text{s}$.