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

Step-by-step 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+(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-(gt^(2))/(2)` (3)

`t=-(2V_(o)sin\theta)/(g)` (4)

Substituting (4) in (1):

`x=-V_(o)cos\theta(2V_(o)sin\theta)/(g)` (5)

`x=-(V_(o)^(2))/(g) sin(2\theta)` (6)

`x=-((9.89 m/s)^(2))/(-9.8m/s^(2)) sin(2(18.7\°))` (7)

Finally:

`x=6.062 m`