A track star in the broad jump goes into the jump at 10 m/s and launches himself at 20° above the horizontal.How long is he in the air before returning to Earth? Note: g = 9.8 m/s². A) 0.35 s B) 0.42 s C) 0.70 s D) 0.89 s

Question

Quizplus · Accepted Answer

We can use the kinematic equation y = yo + voy*t - 0.5*g*t^2 to find the time it takes for the track star to return to the ground.

First, we need to find the initial vertical velocity. We can use the equation voy = vo*sin(theta) = 10*sin(20°) = 3.42 m/s.

Next, we can set y = 0 (since the track star returns to the same height) and solve for t:

0 = 0 + 3.42*t - 0.5*9.8*t^2

This equation simplifies to 4.9t^2 - 3.42t = 0, which factorizes to t(4.9t - 3.42) = 0.

Taking the positive solution, we get t = 0.7 seconds. Therefore, the correct answer is C.

A Track Star in the Broad Jump Goes into the Jump