Final answer:
The student was airborne for approximately 2.3 seconds.
Step-by-step explanation:
First, let's find the time the student was airborne using the horizontal distance:
Use the formula for horizontal distance: d = vt, where d is the distance, v is the horizontal velocity, and t is the time.
Substitute the given values: 20.7 m = v * t.
Next, use the formula for vertical distance: d = 0.5 * g * t^2, where d is the vertical distance, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Substitute the given values: 25.9 m = 0.5 * 9.8 m/s^2 * t^2.
Solve the equation for t: t = sqrt((2 * 25.9 m) / (9.8 m/s^2)).
Calculate the time: t = sqrt(5.28 s^2).
Therefore, the student was airborne for approximately 2.3 seconds.