Marie shoots an arrow vertically upward at a sandhill crane. If the arrow travels to a maximum height of 99.5 meters, calculate how fast it was shot.

And, for how much time was Marie's arrow in the air?

To calculate the initial speed at which Marie's arrow was shot, we can use the equations of motion for vertical motion.

The height of the arrow as a function of time can be described by the equation:
h(t) = (v₀t) - (1/2)gt²

Where:
h(t) is the height of the arrow at time t
v₀ is the initial speed of the arrow
g is the acceleration due to gravity (approximately 9.8 m/s²)
t is the time elapsed

At the maximum height, the velocity of the arrow is zero since it momentarily stops before falling back down. Thus, we can set v(t) = 0 and solve for t.

0 = v₀ - gt
gt = v₀
t = v₀/g

Substituting the given maximum height, h = 99.5 meters, we can solve for v₀.

99.5 = (v₀)(v₀/g)
v₀² = 99.5g
v₀ = √(99.5g)

Using g ≈ 9.8 m/s², we can calculate v₀.

v₀ = √(99.5 * 9.8) ≈ √970.1 ≈ 31.2 m/s

So, Marie shot the arrow with an initial speed of approximately 31.2 m/s.

To find the time the arrow was in the air, we use the equation:

t = v₀/g

t = 31.2/9.8 ≈ 3.18 seconds

Therefore, Marie's arrow was in the air for approximately 3.18 seconds.