Ms. 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 answer these questions, we can use the principles of projectile motion and gravity. Let's break it down step by step:

a) To find the velocity of the arrow just before reaching the ground, we need to find the final velocity of the arrow when it reaches the highest point (95 feet above ground). We can use the equation: v_f = v_i + at, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time taken.

The initial velocity of the arrow can be determined using the equation v_i^2 = u^2 + 2as, where u is the initial velocity, s is the displacement, and a is the acceleration. Since the arrow goes from 1.2m above ground to a height of 95 feet above ground, the displacement will be the difference between these two heights.

Converting the height from feet to meters, the displacement s = (95 - 1.2) * 0.3048. Now we have the displacement and acceleration (due to gravity, which is approximately -9.8 m/s^2). We can solve for the initial velocity:

v_i^2 = u^2 + 2as
0 = u^2 + 2 * (-9.8) * (95 - 1.2) * 0.3048
0 = u^2 - 2 * 9.8 * 93.8 * 0.3048
u^2 = 2 * 9.8 * 93.8 * 0.3048
u ≈ 27.99 m/s (rounded to two decimal places)

Now that we have the initial velocity, we can find the final velocity just before reaching the ground. At its highest point, the velocity will be zero because gravity slows it down and eventually stops it. Therefore, the final velocity is also 0 m/s.

v_f = v_i + at
0 = 27.99 - 9.8t
-27.99 = -9.8t
t ≈ 2.85 seconds (rounded to two decimal places)

(Note: This solution assumes that air resistance is negligible.)

b) Since the arrow penetrates 14 inches into the ground, we can consider it as being decelerated by the Earth and stopping. The acceleration inside the ground can be calculated using the equation: v_f^2 = v_i^2 + 2ad, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and d is the displacement.

Since the arrow comes to a stop, the final velocity is 0 m/s, and the displacement is 14 inches. Converting the displacement to meters, d = 14 * 0.0254. We can solve for the acceleration:

v_f^2 = v_i^2 + 2ad
0 = 27.99^2 + 2 * a * (14 * 0.0254)
0 = 781.52 + 0.7112a
-781.52 = 0.7112a
a ≈ -1098.61 m/s^2 (rounded to two decimal places)

So, the acceleration of the arrow inside the ground is approximately -1098.61 m/s^2.

c) To find the time it takes for the ground to stop the arrow, we can use the equation: v_f = v_i + at.

Since the final velocity just before reaching the ground is 0 m/s, and the initial velocity is 27.99 m/s, we can solve for time:

v_f = v_i + at
0 = 27.99 - 9.8t
-27.99 = -9.8t
t ≈ 2.85 seconds (rounded to two decimal places)

Therefore, the time it takes for the ground to stop the arrow is approximately 2.85 seconds.

d) The direction of the acceleration when the arrow is inside the ground is downward, which is negative in the chosen coordinate system. So, the acceleration is in the negative direction.

Remember to always double-check your calculations and adapt them to any specific conditions or assumptions given in the problem.