An arrow is shot at 28.0 degrees above the horizontal. Its velocity is 52 m/s, and it hits the target.
(a) What is the maximum height the arrow will attain?
(b) The target is at the height from which the arrow was shot. How far away was it?
Vo = 52m/s[28o].
Xo = 52*Cos28 = 45.91 m/s.
Yo = 52*sin28 = 24.41 m/s.
a. Y^2 = Yo^2 + 2g*h.
0 = (24.41)^2 - 19.6h, h = ?.
b. Range = Vo^2*sin(2A)/g.
A = 28o, Range = ?.
To find the maximum height attained by the arrow, we can use the kinematic equations of motion. The key equations we will use are:
1. Vertical motion equation: vf^2 - vi^2 = 2as
2. Velocity equation: vf = vi + at
3. Displacement equation: s = vit + 0.5at^2
Let's start with part (a) - finding the maximum height.
Step 1: Resolve the arrow's initial velocity into horizontal and vertical components.
The initial velocity can be divided into two components: one in the horizontal direction and the other in the vertical direction. Because the angle of the shot is given as 28.0 degrees above the horizontal, we can use trigonometric functions to find these components.
The vertical component (Vy) can be calculated as:
Vy = V * sin(θ)
where V is the magnitude of the initial velocity (52 m/s) and θ is the angle (28.0 degrees).
Step 2: Determine the time taken for the arrow to reach its maximum height.
Once we have the vertical component of the initial velocity, we can calculate the time it takes for the arrow to reach its maximum height. At the maximum height, the vertical velocity component will be zero.
Using the velocity equation (vf = vi + at), we can set the final velocity in the vertical direction (Vfy) to zero and solve for the time (t):
0 = Vy + gt
Rearranging the equation: t = -Vy / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Step 3: Calculate the maximum height (h).
To find the maximum height attained by the arrow, we need to use the displacement equation (s = vit + 0.5at^2). In this case, the displacement is equal to the maximum height (h), the initial velocity is Vy, and the time is t.
Substituting the values into the equation:
h = Vy * t + 0.5 * g * t^2
Now that we have the steps, let's put the values into the equations and calculate the maximum height (h) in part (a).
Vy = V * sin(θ)
t = -Vy / g
h = Vy * t + 0.5 * g * t^2
For part (b) - finding the distance to the target:
Step 1: Calculate the horizontal component of the initial velocity.
We already know the magnitude of the initial velocity (V) and the angle (θ). To find the horizontal component (Vx), we can use the trigonometric functions.
Vx = V * cos(θ)
Step 2: Determine the time of flight.
The time of flight is the total time the arrow remains in the air. Since the vertical motion and horizontal motion are independent of each other (neglecting air resistance), the time of flight (T) only depends on the vertical motion.
Using the formula T = 2t (as the time taken to reach the maximum height and land back on the same level is the same for symmetrical projectile motion).
Step 3: Calculate the distance (d) to the target.
To find the distance, we use the equation: d = Vx * T
Now, let's use these steps to calculate the distance to the target in part (b).