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 solve this problem, we can use the kinematic equations of motion and apply them to different segments of the arrow's motion. Let's break down each part of the problem one by one.
a) The velocity of the arrow just before reaching the ground:
To find the velocity of the arrow just before reaching the ground, we need to consider the vertical motion of the arrow. We can use the equation of motion for vertical displacement (Δy) with constant acceleration (g):
Δy = v₀t + (1/2)gt²
We know the initial vertical displacement Δy = 95ft. The arrow initially starts at a height of 1.2m above the ground, which is equivalent to 3.937ft. Therefore, the total vertical displacement is 95 - 3.937 = 91.063ft.
We also know that the acceleration due to gravity, g, is -32.2 ft/s².
Using these values, we can rearrange the equation and solve for the time it takes for the arrow to reach the ground (t):
91.063 = 0t + (1/2)(-32.2)t²
By solving this quadratic equation, we can find the value of t.
Once we have the value of t, we can use the equation of motion for vertical velocity (v) with constant acceleration (g):
v = v₀ + gt
Substituting the known values, we can calculate the velocity of the arrow just before reaching the ground.
b) The acceleration of the arrow inside the ground:
Inside the ground, the arrow is subjected to an opposing force that causes it to decelerate. Since the arrow penetrates the ground, we can assume constant deceleration.
To find the acceleration, we'll consider the horizontal motion of the arrow. The arrow falls from rest and then penetrates the ground, so its initial horizontal velocity is zero (v₀ = 0).
We'll use the equation of motion for horizontal displacement (Δx) with constant acceleration (a):
Δx = v₀t + (1/2)at²
We know the horizontal displacement (Δx) is 14 inches. Since the horizontal velocity is zero, Δx becomes just the penetration depth into the ground.
Substituting the known values, we can solve for the acceleration (a).
c) The time it takes the ground to stop the arrow:
Once the arrow reaches the ground, it comes to a stop. We need to find the time it takes for the arrow to stop after hitting the ground.
Using the same equation of motion for horizontal displacement (Δx) with constant acceleration (a):
Δx = v₀t + (1/2)at²
In this case, the initial horizontal velocity (v₀) is the same as the velocity just before reaching the ground, which we calculated in part (a). The acceleration (a) is the same as in part (b). The horizontal displacement (Δx) is the penetration depth into the ground.
We can solve for the time (t) it takes for the arrow to come to a stop.
d) The direction of the acceleration:
Since the arrow is decelerating while penetrating the ground, the acceleration will be in the opposite direction of the initial motion, which is upward. Therefore, the direction of the acceleration will be downward.
By following these steps and solving the equations, you can find the answers to each part of the problem.