A pitched ball was hit by a batter at a 30ᵒ angle at a velocity of 40 m/s. The outfield wall, being very short, was just the same height as the swing of the batter’s bat. If the hit was just enough to have been a homerun, how far from the batter was the outfield wall?

Dx = Vo^2*sin(2A)/g.

Dx = 40^2*sin(60)/9.8 = = 141.4 m.

To find the distance from the batter to the outfield wall, we can use the kinematic equations of motion. Let's break down the problem step by step.

1. Decompose the initial velocity into horizontal and vertical components.
Given that the ball was hit at a 30ᵒ angle with a velocity of 40 m/s, we can find the horizontal and vertical components as follows:

Horizontal Component: Vx = V * cosθ
Vertical Component: Vy = V * sinθ

Vx = 40 m/s * cos(30ᵒ) ≈ 34.64 m/s
Vy = 40 m/s * sin(30ᵒ) = 20 m/s

2. Calculate the time taken for the ball to reach the maximum height.
The maximum height occurs when the vertical component of velocity becomes zero.

Using the equation:
Vy = Vy0 + a * t

where initial vertical velocity (Vy0) is 20 m/s and acceleration due to gravity (a) is -9.81 m/s² (negative as it acts against the upward motion),

0 = 20 m/s + (-9.81 m/s²) * t

Solving for t:
t = 20 m/s / 9.81 m/s² ≈ 2.04 s

3. Determine the maximum height reached by the ball.
We can use the equation for motion in the vertical direction:

y = y0 + Vy0 * t + (1/2) * a * t²

Taking y0 = 0 (start at ground level), Vy0 = 20 m/s, a = -9.81 m/s², and t = 2.04 s:

y = 0 + 20 m/s * 2.04 s + (1/2) * (-9.81 m/s²) * (2.04 s)²

y ≈ 20.4 m

4. Calculate the time taken for the ball to hit the ground.
The ball will hit the ground when the vertical component of displacement (y) becomes -20.4 m (negative to return to the ground level).

Using the equation:
y = y0 + Vy0 * t + (1/2) * a * t²

Taking y0 = 0 (start at ground level), Vy0 = 20 m/s, a = -9.81 m/s², and y = -20.4 m:

-20.4 m = 0 + 20 m/s * t - (1/2) * 9.81 m/s² * t²

This results in a quadratic equation which we can solve to find the time it takes for the ball to hit the ground. The positive root will be the solution.
By solving the equation, t ≈ 4.15 s.

5. Calculate the horizontal distance covered by the ball.
The horizontal distance (x) covered by the ball is given by the equation:

x = Vx * t

Taking Vx = 34.64 m/s and t ≈ 4.15 s:

x = 34.64 m/s * 4.15 s ≈ 143.74 m

Therefore, the outfield wall is approximately 143.74 meters away from the batter.