An arrow is shot from 3 ft above the top of a hill with a vertical upward velocity of 108 ​ft/s. If it strikes the plain below after 9.5 ​s, how high is the​ hill?
If the arrow is launched at t​0, then write an equation describing velocity as a function of time?

The height of the hill is approximately 25.73 ft. Where v0 is the initial velocity (108 ft/s), g is the acceleration due to gravity [tex](-32.2 ft/s^2)[/tex],

To find the height of the hill, we can use the formula for the vertical position of an object under constant acceleration:

h = h0 + v0t + 1/2at^2

where h is the final height, h0 is the initial height, v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (-32.2 ft/s^2).

In this case, we are given that the initial height h0 is 3 ft, the initial velocity v0 is 108 ft/s, and the time t is 9.5 s. We want to find the height of the hill, which we can denote as h_hill. The final height is the height of the plain, which we can denote as h_plain and assume is zero.

At the highest point of its trajectory, the arrow will have zero vertical velocity, since it will have stopped rising and just started to fall. So we can set the velocity to zero and solve for the time it takes for that to occur. Using the formula for velocity under constant acceleration:

v = v0 + at

we can solve for t when v = 0, h0 = 3 ft, v0 = 108 ft/s, and a = -32.2 ft/s^2:

0 = 108 - 32.2t

t = 108/32.2 ≈ 3.35 s

Thus, it takes the arrow approximately 3.35 s to reach the top of its trajectory.

Using the formula for the height of an object at a given time, we can find the height of the hill by subtracting the height of the arrow at the top of its trajectory from the initial height:

h_hill = h0 + v0t + 1/2at^2 - h_top

where h_top is the height of the arrow at the top of its trajectory. We can find h_top using the formula for the height of an object at the maximum height of its trajectory:

h_top = h0 + v0^2/2a

Plugging in the given values, we get:

h_top = 3 + (108^2)/(2*(-32.2)) ≈ 196.78 ft

Plugging this into the first equation, we get:

h_hill = 3 + 108(3.35) + 1/2(-32.2)(3.35)^2 - 196.78

h_hill ≈ 25.73 ft

If the arrow is launched at t0, the equation describing velocity as a function of time would be:

v(t) = v0 - gt

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