A batted baseball leaves the bat with an angle of 30* above the horizontal is caught by an outfielder 120 m from the plate.
a. What was the initial velocity of the ball?
b. How high did it rise?
c. How long wasit in the air?
(a) to find v, use the range v^2/2g sin2θ = 120
(b) h = v^2/2g sin^2 θ
(c) v-gt = 0 gives the rise time. Double that to find total flight time
To solve this problem, we can use the following equations of motion:
1. Vertical motion equation (y-direction):
y = yo + vot + 0.5at^2
2. Horizontal motion equation (x-direction):
x = xo + vot
Given:
Angle of ball leaving the bat (θ) = 30 degrees
Distance to the outfielder (x) = 120 m
Acceleration due to gravity (g) = 9.8 m/s^2
Let's solve this step by step:
a. What was the initial velocity of the ball?
To find the initial velocity of the ball, we need to separate the velocity vector into its horizontal and vertical components.
Vertical component:
voy = v * sin(θ)
Horizontal component:
vox = v * cos(θ)
Since the horizontal component of velocity remains constant throughout the motion, we can use the horizontal equation to find the initial velocity (v).
From the horizontal equation:
120 m = 0 + v * cos(30)
Solving for v:
v = 120 m / cos(30)
v ≈ 138.56 m/s
Therefore, the initial velocity of the ball is approximately 138.56 m/s.
b. How high did it rise?
To find the maximum height or how high the ball rises, we need to find the vertical displacement (Δy) using the vertical motion equation.
Let's assume the initial height (yo) is 0.
Using the vertical equation:
0 = 0 + v * sin(30) * t - 0.5 * g * t^2
Since the ball reaches its maximum height when the vertical velocity (v * sin(30)) becomes 0, we can express the time (t) in terms of the initial velocity (v) and acceleration due to gravity (g):
t = (v * sin(30)) / g
Substituting this value back into the vertical motion equation, we can find the vertical displacement (Δy).
Δy = 0 + v * sin(30) * [(v * sin(30)) / g] - 0.5 * g * [(v * sin(30)) / g]^2
Simplifying:
Δy = (v^2 * sin^2(30)) / (2 * g)
Substituting the known values:
Δy = (138.56^2 * sin^2(30)) / (2 * 9.8)
Δy ≈ 51.03 m
Therefore, the ball rises to approximately 51.03 m.
c. How long was it in the air?
To find the total time the ball was in the air, we can use the formula for the time of flight:
t = (2 * v * sin(30)) / g
Substituting the known values:
t = (2 * 138.56 * sin(30)) / 9.8
t ≈ 8.88 s
Therefore, the ball was in the air for approximately 8.88 seconds.
To solve this problem, we can use the principles of projectile motion. We will need to break the initial velocity into its horizontal and vertical components. Let's start with part (a) of the question.
a. What was the initial velocity of the ball?
To find the initial velocity of the ball, we need to first find its vertical and horizontal component velocities. We know that the ball is hit with an angle of 30 degrees above the horizontal.
The horizontal component of velocity (Vx) remains constant throughout the motion, and the vertical component of velocity (Vy) changes due to the effect of gravity.
Vy = V * sin(θ)
Vx = V * cos(θ)
Where:
- V is the initial velocity of the ball
- θ is the angle of elevation (30 degrees in this case)
- Vy is the vertical component of velocity
- Vx is the horizontal component of velocity
Since we have the angle of elevation and the vertical component of velocity, we can solve for the initial velocity, V:
V = Vy / sin(θ)
Given that the ball is caught 120 m away from the plate and neglecting air resistance, we can assume that the horizontal distance traveled (range) is equal to the horizontal component of velocity multiplied by the time of flight:
Range = Vx * time
We can write the time of flight, t, as:
time = Range / Vx
Now, let's move on to part (b) of the question.
b. How high did it rise?
To find the maximum height reached by the ball, we need to determine the time it takes for the ball to reach its maximum height. At the maximum height, the vertical component of velocity becomes zero (Vy = 0), so we can find the time using the equation:
Vy = Vy0 - gt
where:
- Vy is the vertical component of velocity at any point during the motion
- Vy0 is the initial vertical component of velocity (Vy0 = V * sin(θ))
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time taken to reach the maximum height
Solving the equation for t, we get:
t = Vy0 / g
Once we have the time, we can find the maximum height (H) using the equation:
H = (Vy0^2) / (2 * g)
Finally, let's move on to part (c) of the question.
c. How long was it in the air?
The time of flight, which is the total time the ball is in the air, can be calculated using the equation:
total time = 2 * t
Now that we have the explanations, let's calculate the solutions using the given information.