Mr. Anderson shoots an arrow straight up into the air. The arrow leaves her

bow at 1.2[m] above ground, and reaches a height of 95 feet above ground.
It then falls down and penetrates 14 inches into the ground. Find:
a) The velocity of the arrow just before reaching the ground.
b) The acceleration of the arrow inside the ground (assume constant).
c) The time it takes the ground to stop the arrow.
d) The direction of the acceleration.

To solve the problem, we need to use the equations of motion for vertically launched projectiles.

a) To find the velocity of the arrow just before reaching the ground, we can use the equation:

v^2 = u^2 + 2as

Where:
v = final velocity (what we want to find)
u = initial velocity (when the arrow leaves the bow)
a = acceleration (which is in this case, the acceleration due to gravity. We assume it to be -9.81 m/s^2 as it acts in the opposite direction of the arrow's motion)
s = displacement (which is the distance the arrow has fallen, in this case, the height the arrow reached above the ground)

The initial velocity u can be found using the equation of motion:

v = u + at

Since the arrow is launched vertically, its initial velocity u is equal to the velocity just before it reaches the ground. We can rearrange this equation to solve for u:

u = v - at

b) The acceleration of the arrow inside the ground is given as "assume constant." This means we will assume the arrow experiences the same acceleration while penetrating the ground as it does in free fall, which is the acceleration due to gravity, -9.81 m/s^2.

c) To find the time it takes for the ground to stop the arrow, we need to determine the time it takes for the arrow to fall from its maximum height to the ground. We can use the equation:

s = ut + (1/2)at^2

Rearranging this equation, we have:

t = sqrt(2s/a)

Where:
t = time
s = displacement (the height the arrow reached above the ground)
a = acceleration due to gravity

d) Since the acceleration due to gravity is acting in the opposite direction of the arrow's motion, the direction of the acceleration inside the ground is downward.

Now, we can substitute the given values and solve for the required answers.