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

Answer:

x = 9.5 m

Step-by-step 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

$t = \frac{-4.10}{-9.8}$

$t = 0.418 s$

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

to find horizontal distance covered:

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