If a jumper has a vertical leap of 1.54 m on his trial run, what is their takeoff speed and hang time (total time to move upwards to the peak and then return to the ground)?

In this scenario, the takeoff speed is the same as the initial velocity of the jumper and it can be calculated by using the third equation of motion. Also, since the motion is against gravity, the acceleration due to gravity will be negative (g = -9.81 m/s²)

V² = U² - 2gS

U² = V² + 2gS

U² = 0² + 2(9.81)(1.54)

U = √30.22

U = 5.50 m/s.

Next, we would calculate the peak time by using the first equation of motion:

V = U - at

0 = 5.5 - 9.81t

9.81t = 5.5

t = 5.5/9.81

t = 0.56 seconds.

Mathematically, the hang time is double the peak time:

Hang time = 2 × 0.56

Hang time = 1.12 seconds.

Read more on hang time here: https://brainly.com/question/14902568

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If a jumper has a vertical leap of 1.54 m on his trial run, what is their takeoff speed and hang time (total time to move upwards to the peak and then return to the ground)?​

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How to calculate takeoff speed and hang time?

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If a jumper has a vertical leap of 1.54 m on his trial run, what is their takeoff speed and hang time (total time to move upwards to the peak and then return to the ground)?