Question

A long jumper leaves the ground at an angle of 18.7 ◦ to the horizontal and at a speed of 9.89 m/s. How far does he jump? The acceleration due to gravity is 9.8 m/s 2 .

Answer 1

Answer: 6.062 m

Explanation:

This situation is related to projectile motion or parabolic motion, which has two components: x-component and y-component. Being their main equations as follows:  

x-component:  

$x=V_{o}cos\theta t$   (1)  

Where:  

$V_{o}=9.89 m/s$ is the jumper's initial velocity

$\theta=18.7\°$ is the angle  

$t$ is the time since the jumperleaves the ground and falls again

y-component:  

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$   (2)  

Where:  

$y_{o}=0 m$  is the initial height of the jumper

$y=0 m$  is the final height of the jumper (when it finally hits the ground)  

$g=-9.8m/s^{2}$  is the acceleration due gravity

We need to find the horizontal distance, but firstly we will find the time with (2):

$0 m=0 m+V_{o}sin\theta t-\frac{gt^{2}}{2}$   (3)  

$t=-\frac{2V_{o}sin\theta}{g}$   (4)  

Substituting (4) in (1):

$x=-V_{o}cos\theta\frac{2V_{o}sin\theta}{g}$   (5)  

$x=-\frac{V_{o}^{2}}{g} sin(2\theta)$   (6)  

$x=-\frac{(9.89 m/s)^{2}}{-9.8m/s^{2}} sin(2(18.7\°))$   (7)  

Finally:

$x=6.062 m$