Understanding the Concepts

* Constant Acceleration: The arrow is under the influence of gravity, which provides a constant downward acceleration (approximately 9.8 m/s²).

* Kinematic Equations: We can use the following kinematic equation to relate displacement, initial velocity, acceleration, and time:

* d = v₀t + (1/2)at²

* where:

* d = displacement (75 m)

* v₀ = initial velocity (what we want to find)

* t = time in the air (also what we want to find)

* a = acceleration due to gravity (-9.8 m/s²)

Calculations

1. Finding the Time to Reach Maximum Height:

* At the maximum height, the arrow's velocity is 0 m/s.

* We can use the following equation to find the time it takes to reach this point:

* v = v₀ + at

* 0 = v₀ + (-9.8)t

* v₀ = 9.8t

2. Finding the Initial Velocity:

* Since the arrow travels up and then back down, the total time in the air is twice the time it takes to reach the maximum height.

* Let's call the time to reach the maximum height 't'. The total time in the air is '2t'.

* We can now use the first kinematic equation:

* d = v₀t + (1/2)at²

* 75 = v₀t + (1/2)(-9.8)(2t)²

* 75 = v₀t - 19.6t²

* Substitute v₀ = 9.8t from step 1:

* 75 = (9.8t)t - 19.6t²

* 75 = 9.8t² - 19.6t²

* 75 = -9.8t²

* t² = -75 / -9.8 ≈ 7.65

* t ≈ √7.65 ≈ 2.77 seconds (This is the time to reach the maximum height)

3. Calculating Initial Velocity:

* Use the equation v₀ = 9.8t:

* v₀ = 9.8 * 2.77 ≈ 27.2 m/s

Answers

* Initial velocity: The arrow left the bow with a velocity of approximately 27.2 m/s.

* Time in the air: The arrow was in the air for approximately 5.54 seconds (2 * 2.77 seconds).