Question

A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 20° above the horizontal. How far does the jumper jump?

Answer 1

x = 9.5 m

Explanation:

Given:

This problem is based on 2D motion kinematics equations:

`V_0 = 12 m/s`

`\theta = 20^0`

`g=9.8 m/s^2`

At maximum height:

`V_y = 0`

Components of initial velocity:

`V_0x = V_0 cos (\theta) =(12 )(cos (20^0)) = 11.3 m/s`

`V_0x = V_0 sin (\theta) = (12)(sin (20^0)) = 4.10 m/s`

To find time to reach maximum height:

`V_y = V_0y -gt`

Plugging in the known values:

0 = 4.10 - 9.8t

`t = (-4.10)/(-9.8)`

`t = 0.418 s`

Total time = 2t = 2(0.418) = 0.836 s

to find horizontal distance covered:

`x = (V_0x)(t)\\x = (11.3)(0.836)\\x = 9.5 m`