A batter hits a fly ball which leaves the bat 0.90 m above the ground at an angle of 61 degrees with an initial speed of 28 m/s heading towards center field. Ignore air resistance. (a) How far from the home plate would the ball land if not caught? (b) The ball is caught by the centerfielder who, starting at a distance of 105m from the home plate, runs straight towards home plate at a constant speed and makes the catch at ground level. What was his speed?
divide into vertical and horizontal problems
Vertical problem:
initial vertical speed = Vi = 28 sin 61
calculate Vi = 24.5 m/s
Now
v = Vi - 9.81 t
h = Hi + Vi t - 4.9 t^2 where Hi is initial height of 0.90 meter
h = 0 at ground
so
4.9 t^2 - Vi t - 0.90= 0
4.9 t^2 - 24.5 t - 0.90 = 0
solve quadratic
https://www.mathsisfun.com/quadratic-equation-solver.html
t = 5.04 seconds to hit ground
Now horizontal problem
u = 28 cos 61
distance = (28 cos 61) * 5.04
fielder goes (105 - distance)meters in 5.04 seconds
a
(a) Ah, the age-old question of baseball and projectile motion. Let's calculate! Since we're ignoring air resistance, we can treat the horizontal and vertical motion independently.
To find the horizontal distance the ball would travel, we first need to calculate the time it takes to reach the ground. We can use the vertical motion equation:
h = vit + (1/2)at^2
In this case, the initial vertical velocity (vi) is the vertical component of the initial velocity (28 m/s) times the sine of the launch angle (61 degrees).
vi = 28 m/s * sin(61 degrees)
Next, we have the acceleration due to gravity (a), which is -9.8 m/s^2 because it acts downward.
Using the equation, h = 0.9 m, and solving for t, we get:
0.9 m = (28 m/s * sin(61 degrees))t + (0.5)(-9.8 m/s^2)t^2
Solve the equation for t, and you'll find the time it takes for the ball to reach the ground.
Now we can calculate the horizontal distance, x. We can use the horizontal motion equation:
x = vix * t
Since we're ignoring air resistance, the horizontal velocity (vix) remains constant and equal to the initial velocity (28 m/s) times the cosine of the launch angle (61 degrees).
x = (28 m/s * cos(61 degrees)) * t
Plug in the value of t you found, and you'll have your answer for (a)!
(b) Now, let's tackle the catcher's speed. Since the centerfielder runs straight towards home plate, he intercepts the ball on the path perpendicular to the initial motion of the ball.
We know the centerfielder starts at a distance of 105m from the home plate and intercepts the ball at ground level. The horizontal distance covered by the centerfielder will be equal to the horizontal distance traveled by the ball.
Using the same formula as before, x = (28 m/s * cos(61 degrees)) * t, we can calculate the time it takes for the centerfielder to reach the ball.
Finally, the speed of the centerfielder can be calculated by dividing the total distance traveled (105m) by the time it took him to reach the ball.
So there you have it, my friend! Use these calculations to find the answers you seek. Good luck, and may the baseball gods be with you!
To solve this problem, we can break it down into two parts: (a) determining the horizontal distance the ball would travel if not caught, and (b) finding the speed of the centerfielder.
Step 1: Determining the horizontal distance traveled by the ball if not caught.
Let's solve part (a) first. We can use the equations of projectile motion to determine the horizontal distance traveled by the ball.
Given:
- Initial speed (V₀) = 28 m/s
- Launch angle (θ) = 61 degrees
- Initial height (y) = 0.90 m
We can decompose the initial velocity into its horizontal and vertical components:
V₀x = V₀ * cos(θ)
V₀y = V₀ * sin(θ)
Horizontal distance traveled (x) can be calculated using the equation:
x = V₀x * t
To find the time it takes for the ball to land, we need to determine the time it takes to reach the maximum height (t₁) and the total time of flight (t). We can use the equation:
- Maximum height time (t₁) = (V₀y) / (g)
- Total time of flight (t) = 2 * t₁
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Finally, substitute the time (t) into the equation for horizontal distance (x) to find the answer to part (a).
Step 2: Finding the speed of the centerfielder.
To solve part (b), we need to determine the speed of the centerfielder. We know:
- Distance covered by the centerfielder (d) = 105 m
- Horizontal distance traveled by the ball if not caught (from part (a))
We can calculate the time (t₂) it took for the centerfielder to reach the home plate using the equation:
t₂ = d / x
Finally, the speed of the centerfielder can be calculated using the equation:
Speed = d / t₂
Now, let's calculate the answers step by step:
Step 1: Determining the horizontal distance traveled by the ball if not caught.
1. Calculate V₀x:
V₀x = V₀ * cos(θ)
V₀x = 28 m/s * cos(61°)
2. Calculate V₀y:
V₀y = V₀ * sin(θ)
V₀y = 28 m/s * sin(61°)
3. Calculate the maximum height time (t₁):
t₁ = V₀y / g
4. Calculate the total time of flight (t):
t = 2 * t₁
5. Calculate the horizontal distance traveled (x):
x = V₀x * t
Step 2: Finding the speed of the centerfielder.
6. Calculate the time taken by the centerfielder to reach the home plate (t₂):
t₂ = d / x
7. Calculate the speed of the centerfielder:
Speed = d / t₂
Let's perform the calculations.
To solve this problem, we can break it down into two parts:
(a) calculating the horizontal distance the ball would travel if not caught and
(b) calculating the speed of the centerfielder.
(a) To determine how far the ball would land from home plate if it were not caught, we need to find the horizontal distance covered by the ball. We can do this by using the initial velocity, launch angle, and neglecting air resistance.
Step 1: Break the initial velocity into vertical and horizontal components.
The vertical component, Vy, can be found using trigonometry:
Vy = V * sinθ
Vy = 28 m/s * sin(61 degrees) ≈ 24.01 m/s
The horizontal component, Vx, can also be found using trigonometry:
Vx = V * cosθ
Vx = 28 m/s * cos(61 degrees) ≈ 13.45 m/s
Step 2: Determine the time of flight.
To calculate the time it takes for the ball to land, we can use the equation:
y = Vyt - 1/2gt^2
Since the ball lands at the same height it was launched from, y = 0.90 m.
0 = 24.01 m/s * t - 1/2 * 9.8 m/s^2 * t^2
0 = 24.01t - 4.9t^2
4.9t^2 - 24.01t = 0
t(4.9t - 24.01) = 0
Solving for t, we find two solutions: t = 0 (the initial time) and t ≈ 4.90 s.
Step 3: Calculate the horizontal distance.
The horizontal distance covered by the ball is given by:
distance = Vx * time
distance = 13.45 m/s * 4.90 s ≈ 65.8 m
Therefore, the ball would land approximately 65.8 meters from home plate if it were not caught.
(b) To determine the speed of the centerfielder, we need to find the time it takes for the centerfielder to reach the ball. Since he starts at a distance of 105 m from home plate and runs straight towards it, we can use the equation:
distance = speed * time
The distance is 105 m, and the time is the same as the time calculated in part (a), which is approximately 4.90 s.
105 m = speed * 4.90 s
Solving for speed, we get:
speed = 105 m / 4.90 s ≈ 21.43 m/s
Therefore, the centerfielder's speed when making the catch is approximately 21.43 m/s.