Question

Mayan kings and many school sports teams are named for the puma, cougar or mountain lionFelis conclude the best jumper among animals. It can jump to a height of $12.0ft$ when leaving the ground at an angle of ${45.0^0}$ with what speed, in SI units, does it leave the ground to make this leap?

Answer 1

In order to solve this question, we will resolve the components of velocity in X and Y direction and then by using Newton’s equation of motion which is ${v^2} - {u^2} = 2aS$ where $v,u$ are the final and initial velocity of the body with acceleration a and covering distance of S. In case of body moving against gravity acceleration is taken as acceleration due to gravity with negative sign.

Complete step by step answer:
According to the question, when jumper leaves the ground at an angle of ${45.0^0}$ , let $v$ be the net magnitude of velocity of jumper at initial point and when it reaches the highest point at a height of $12.0ft$ its velocity will became zero.

So, the initial velocity of the jumper is $v$.Final velocity of the jumper at the highest point is $0$. Acceleration due to gravity is $- g = - 9.8\,m{s^{ - 2}}$. And distance covered in this journey is $S = 12ft = 3.66\,m$ now using the formula,
${v^2} - {u^2} = 2aS$ Put the values of each parameters we get,
$\Rightarrow 0 - {v^2} = - 2gS$
$\Rightarrow v = \sqrt {2 \times 9.8 \times 3.66}$
$\Rightarrow v = 8.47\,m{s^{ - 1}}$

Now the net magnitude of velocity of the jumper is $v = 8.47\,m{s^{ - 1}}$.But since, jumper covers the journey in Y direction which is height so only normal component of velocity will be needed so if ${v_{launch}}$ is the needed velocity to jump then,
${v_{launch}}\sin {45^0} = v$
Putting the values we get,
${v_{launch}} = 8.47 \times \sqrt 2$
$\therefore {v_{launch}} = 12.0\,m{s^{ - 1}}$

Hence, the velocity needed to jump is ${v_{launch}} = 12.0\,m{s^{ - 1}}$.

Note: It should be remembered that, change the distance of $12.0\,ft$ in to meter with conversion as $1\,ft = 0.3048\,m$ and the value of trigonometric function $\sin {45^0} = \dfrac{1}{{\sqrt 2 }}$ since, jumper covers the height which is in Y direction in XY plane so we consider only normal component of velocity while calculating launch speed.