Question
An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal. How long does the athlete stay in the air, assuming no significant air resistance?
Answers
oobleck
he is halfway through his jump when he reaches the top. That is when the vertical speed has gone to zero. That will be when
9.14 sin35° - 9.81t = 0
t = 9.14 sin35°/9.81
So he will be in the air for 2t seconds by the time he is back on the ground.
9.14 sin35° - 9.81t = 0
t = 9.14 sin35°/9.81
So he will be in the air for 2t seconds by the time he is back on the ground.