A baseball leaves the bat with a speed of 44.0 m/s and an angle of 30.0 above the horizontal. A 10.0-m-high fence is located at a horizontal distance of 132 m from the point where the ball is struck. Assuming the ball leaves the bat 1.0 m above ground level, by how much does the ball clear the fence?
A) 4.4 m
B) 8.4 m
C) 13.4 m
D) 17.4 m
E) 18.4 m
Well, let's see here. A baseball leaves the bat with a speed of 44.0 m/s and an angle of 30.0 degrees above the horizontal. It's like the ball is saying, "I'm going up, up, and away!" But then it realizes there's a 10.0-m-high fence in its way. Talk about a buzzkill.
So, how much does the ball clear the fence? To find out, we need to calculate the horizontal and vertical components of the ball's initial velocity.
First, let's break down the initial velocity. The horizontal component is given by Vx = V * cos(theta), where V is the speed and theta is the angle above the horizontal. So, Vx = 44.0 m/s * cos(30.0 degrees) = 38.08 m/s.
The vertical component is given by Vy = V * sin(theta), where V is the speed and theta is the angle above the horizontal. So, Vy = 44.0 m/s * sin(30.0 degrees) = 22.0 m/s.
Now, let's calculate the time it takes for the ball to reach the fence. We can use the equation d = Vx * t, where d is the horizontal distance and t is the time. Rearranging the equation to solve for t, we get t = d / Vx. Plugging in the values, we have t = 132 m / 38.08 m/s = 3.46 s.
Next, we can calculate how high the ball reaches at that time using the equation Y = Y0 + Vy*t - (1/2) * g * t^2, where Y is the vertical position, Y0 is the initial vertical position, Vy is the vertical velocity, t is the time, and g is the acceleration due to gravity. Plugging in the values, we have Y = 1.0 m + (22.0 m/s)(3.46 s) - (1/2)(9.8 m/s^2)(3.46 s)^2 = 37.9 m.
Finally, we can calculate how much the ball clears the fence by subtracting the height of the fence from the maximum height the ball reaches. So, the ball clears the fence by 37.9 m - 10.0 m = 27.9 m.
And the answer is... drumroll, please... E) 27.9 m! Way to go, ball! You really cleared that fence with style!
To find out how much the ball clears the fence, we need to determine the maximum height the ball reaches and the horizontal distance it travels before hitting the ground.
1. Find the time it takes for the ball to reach its maximum height:
- The initial vertical velocity of the ball is given by v₀y = v₀ * sin(θ), where v₀ = 44.0 m/s and θ = 30.0°.
v₀y = 44.0 m/s * sin(30.0°) ≈ 22.0 m/s.
- The acceleration due to gravity is a = -9.8 m/s².
- Use the kinematic equation: v = v₀ + a * t, where v is the final velocity (0 m/s) at the maximum height.
0 m/s = 22.0 m/s + (-9.8 m/s²) * t.
- Solve for t:
-9.8 m/s² * t = -22.0 m/s.
t ≈ 2.24 s.
2. Find the maximum height reached by the ball:
- Use the kinematic equation: θ = v₀y * t + (1/2) * a * t².
θ = 22.0 m/s * 2.24 s + (1/2) * (-9.8 m/s²) * (2.24 s)².
- Solve for θ:
θ ≈ 24.8 m.
3. Find the horizontal distance traveled by the ball:
- The initial horizontal velocity of the ball is given by v₀x = v₀ * cos(θ), where v₀ = 44.0 m/s and θ = 30.0°.
v₀x = 44.0 m/s * cos(30.0°) ≈ 38.1 m/s.
- Use the kinematic equation: d = v₀x * t, where d is the horizontal distance traveled.
d = 38.1 m/s * 2.24 s ≈ 85.3 m.
4. Subtract the horizontal distance from the total horizontal distance to the fence to find out how much the ball clears the fence:
Total horizontal distance = 132 m.
Clearance = Total horizontal distance - Horizontal distance traveled.
Clearance = 132 m - 85.3 m ≈ 46.7 m.
Therefore, the ball clears the fence by approximately 46.7 meters. So none of the given answer choices are correct.
To solve this problem, we can break the initial velocity of the baseball into horizontal and vertical components.
The horizontal component of velocity remains constant throughout the motion and does not affect the vertical motion. The initial horizontal velocity (Vx) can be found using the equation:
Vx = V * cos(θ)
where V is the initial speed of the ball (44.0 m/s) and θ is the angle above the horizontal (30.0°). Plugging in the values:
Vx = 44.0 m/s * cos(30.0°)
Vx = 44.0 m/s * (√3/2)
Vx ≈ 38.1 m/s
The vertical component of velocity (Vy) changes due to the gravitational acceleration during the motion of the ball. The initial vertical velocity (Vy0) can be found using the equation:
Vy0 = V * sin(θ)
where V is the initial speed of the ball (44.0 m/s) and θ is the angle above the horizontal (30.0°). Plugging in the values:
Vy0 = 44.0 m/s * sin(30.0°)
Vy0 = 44.0 m/s * (1/2)
Vy0 = 22.0 m/s
To find the time it takes for the ball to reach the fence, we can use the equation for vertical displacement:
Δy = Vy0 * t + (1/2) * a * t^2
where Δy is the vertical displacement (10.0 m), Vy0 is the initial vertical velocity (22.0 m/s), t is the time, and a is the acceleration due to gravity (-9.8 m/s^2).
Plugging in the values, we get:
10.0 m = 22.0 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
Rearranging the equation, we obtain a quadratic equation:
(1/2) * (-9.8 m/s^2) * t^2 + 22.0 m/s * t - 10.0 m = 0
Solving this quadratic equation gives us the time it takes for the ball to reach the fence.
Once we have the time, we can find the horizontal displacement (Δx) using the equation:
Δx = Vx * t
where Vx is the initial horizontal velocity (38.1 m/s) and t is the time.
By subtracting the horizontal distance from the point where the ball is struck (132 m) from the horizontal displacement, we can find by how much the ball clears the fence.
Now, let's calculate the answer.
The travel time to the fence is
T = 132 m/(Vo*cos 30) = 3.464 s
Vo = 44 m/s.
Write an equation for the height (y) of the ball vs time and determine its height at time T.
y = 1 + Vo*sin 30*T - (g/2)*T^2