Question

A baseball player hit a ball from the ground with a speed of 30 m/s and an angle of 36.9 above the horizontal. Ignore air resistance. What is the ball's velocity when it returns to the ground? when is the ball 10 m above the ground?

Answer 1

Kindly upvote:)

Answer 2

To solve this problem, we can break it into two parts: finding the ball's velocity when it returns to the ground and finding the time when the ball is 10 meters above the ground.

1. Finding the Ball's Velocity when it Returns to the Ground:

You can use the following equations to find the horizontal and vertical components of the ball's velocity:

Horizontal component of velocity (Vx): Vx = V₀ * cos(θ)

Vertical component of velocity (Vy): Vy = V₀ * sin(θ)

Where:

• V₀ is the initial velocity (30 m/s).

• θ is the launch angle (36.9 degrees, which must be converted to radians).

First, convert the angle to radians: θ (radians) = 36.9 degrees * (π/180) ≈ 0.6435 radians

Now, calculate the horizontal and vertical components of velocity:

Vx = 30 m/s * cos(0.6435) ≈ 30 m/s * 0.795 ≈ 23.85 m/s Vy = 30 m/s * sin(0.6435) ≈ 30 m/s * 0.606 ≈ 18.18 m/s

Now, to find the ball's velocity when it returns to the ground, consider the vertical motion. The ball goes up until it reaches its maximum height, then it falls back down to the ground. The time to reach the maximum height can be found using the vertical component of velocity:

Vy = 18.18 m/s (initial vertical velocity) g = 9.81 m/s² (acceleration due to gravity, acting downward)

Use the following equation to find the time to reach the maximum height (t_max):

t_max = Vy / g ≈ 18.18 m/s / 9.81 m/s² ≈ 1.853 seconds

The total time for the ball to return to the ground is twice this time (up and down):

Total time = 2 * t_max ≈ 2 * 1.853 seconds ≈ 3.706 seconds

Now, calculate the horizontal distance traveled by the ball during this time:

Horizontal distance = Vx * Total time ≈ 23.85 m/s * 3.706 seconds ≈ 88.16 meters

So, the ball's velocity when it returns to the ground is approximately 23.85 m/s horizontally.

2. Finding When the Ball is 10 Meters Above the Ground:

You can use the following kinematic equation to find the time (t) when the ball is 10 meters above the ground:

y = y₀ + V₀y * t - (1/2) * g * t²

Where:

• y₀ is the initial vertical position (0 meters from the ground).

• V₀y is the initial vertical velocity (18.18 m/s).

• g is the acceleration due to gravity (9.81 m/s²).

• y is the final vertical position (10 meters above the ground).

Now, plug in the values:

10 m = 0 m + 18.18 m/s * t - (1/2) * 9.81 m/s² * t²

Rearrange the equation:

(1/2) * 9.81 m/s² * t² - 18.18 m/s * t + 10 m = 0

You can solve this quadratic equation for t using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

Where: a = (1/2) * 9.81 m/s² b = -18.18 m/s c = 10 m

t = [18.18 ± √((-18.18)² - 4 * (1/2) * 9.81 * 10)] / (2 * (1/2) * 9.81)

Calculate the two possible values of t, but you'll only consider the positive value since you're interested in when the ball is 10 meters above the ground:

t ≈ 3.499 seconds

So, the ball is approximately 10 meters above the ground at t ≈ 3.499 seconds.