# A mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m. What is the speed of the mountain lion just as it leaves the ground?

The aim of this question is to utilize the equations of motion for solving 2D motion-related problems.

The speed is the rate of change of distance s with respect to time t:

v = s/t

If vf is the final speed, vi is the initial speed, a is the acceleration and s is the distance covered, the equations of motion are given by:

$v_{ f } \ = \ v_{ i } + a t$

$S = v_{i} t + \dfrac{ 1 }{ 2 } a t^2$

$v_{ f }^2 \ = \ v_{ i }^2 + 2 a S$

For vertical upward motion:

$v_{ fy } \ = \ 0, \ and \ a \ = \ -9.8$

For vertical downward motion:

$v_{ iy } \ = \ 0, \ and \ a \ = \ 9.8$

We will use a combination of the above constraints and equations to solve the given problem.

Using the 3rd equation of motion in the vertical direction:

$v_{ fy }^2 \ = \ v_{ iy }^2 + 2 a S$

Substituting values:

$( 0 )^2 \ = \ v_{ iy }^2 + 2 ( -9.8 ) ( 3 )$

$\Rightarrow 0 \ = \ v_{ iy }^2 \ – \ 58.8$

$\Rightarrow v_{ iy }^2 \ = \ 58.8$

$\Rightarrow v_{ iy } \ = \ \sqrt{ 58.8 }$

$\Rightarrow v_{ iy } \ = \ 7.668 m/s$

Using second equation of motion:

$S = v_{iy} t + \dfrac{ 1 }{ 2 } a t^2$

Substituting values:

$3 \ = \ ( 0 ) t + \dfrac{ 1 }{ 2 } (9.8) t^2$

$\Rightarrow 3 \ = \ 4.9 t^2$

$\Rightarrow t \ = \ \sqrt{ \dfrac{ 3 }{ 4.9 } }$

$\Rightarrow t \ = \ 0.782 \ s$

Using the formula for speed in horizontal direction:

$v_x \ = \ \dfrac{ 10 }{ 0.782 } = 12.78 \ m/s$

Calculating the magnitude of the speed:

$|v| \ = \ \sqrt{ v_x^2 \ + \ v_y^2 }$

$\Rightarrow |v| \ = \ \sqrt{ ( 12.78 )^2 \ + \ ( 7.668 )^2 }$

$\Rightarrow |v| \ = \ 14.9 \ m/s$

Calculating the direction of the speed:

$\theta \ = \ tan^{-1} \bigg ( \dfrac{ v_y }{ v_x } \bigg )$

$\theta \ = \ 36.9^{ \circ }$

## Numerical Result

$v \ = \ 14.9 \ m/s \text{ at } \theta = 36.9^{ \circ } \text{ from ground }$

## Example

A man makes a leap $2.0 \ m$ long and $0.5 \ m$ high. What is the speed of the man just as he leaves the ground?

Using the 3rd equation of motion in the vertical direction:

$v_{ fy }^2 \ = \ v_{ iy }^2 + 2 a S$

$\Rightarrow v_{ iy } \ = \ \sqrt{ -2 a S – v_{ fy }^2 }$

$\Rightarrow v_{ iy } \ = \ \sqrt{ -2 ( -9.8 ) ( 0.5 ) – 0 } \ = \ 9.8 \ m/s$

Using second equation of motion:

$S = v_{iy} t + \dfrac{ 1 }{ 2 } a t^2$

$0.5 \ = \ ( 0 ) t + \dfrac{ 1 }{ 2 } (9.8) t^2$

$\Rightarrow t \ = \ \sqrt{ \dfrac{ 0.5 }{ 4.9 } } \ = \ 0.32 \ s$

Using the formula for speed in horizontal direction:

$v_x \ = \ \dfrac{ 2 }{ 0.32 } = 6.25 \ m/s$

Calculating the magnitude of the speed:

$|v| \ = \ \sqrt{ v_x^2 \ + \ v_y^2 } \ = \ \sqrt{ ( 6.25 )^2 \ + \ ( 9.8 )^2 } \ = \ 11.62 \ m/s$

Calculating the direction of the speed:

$\theta \ = \ tan^{-1} \bigg ( \dfrac{ v_y }{ v_x } \bigg ) \ = \ tan^{-1} \bigg ( \dfrac{ 9.8 }{ 6.25 } \bigg ) \ = \ 57.47^{ \circ }$