A track star in the broad jump goes into the jump at 12 m/s and launches himself at 20° above the horizontal.How long is he in the air before returning to Earth? (g = 9.8 m/s²)

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the kinematic equations of motion for projectile motion. First, we need to find the vertical and horizontal components of the velocity:

Vertical component: 
vy = vi*sinθ = 12*sin20° = 4.1 m/s

Horizontal component: 
vx = vi*cosθ = 12*cos20° = 11.1 m/s

Next, we can use the formula to find the time of flight:
time of flight = (2*vy)/g

time of flight = (2*4.1)/9.8 ≈ 0.84 seconds
Therefore, the track star is in the air for 0.84 seconds before returning to Earth. The best choice is B.

A Track Star in the Broad Jump Goes into the Jump