A baseball is struck by a bat and 3 second later it is caught 30 ma way. If it is one meter above the ground when struck and caught, find the greatest height it reached above the ground.
It was in the air for 3 seconds, so it went up for 1.5 seconds
v = Vi - g t
0 at top = Vi - 9.8 (1.5)
so Vi, initial up speed = 14.7 m/s
H = Hi + Vi t - 4.9 t^2
H = 1 + 14.7(1.5) - 4.9(1.5)^2
H = 12.025 m
thank you ! God bless you! :)
To find the greatest height the baseball reached above the ground, we can use the equations of motion. Let's break down the problem step by step:
Step 1: Define the given variables:
- Time of flight (t): 3 seconds
- Horizontal distance (d): 30 m
- Initial vertical position (y₀): 1 m (above the ground)
- Final vertical position (y): 1 m (same as initial position)
Step 2: Find the vertical velocity (Vy) at the time of flight using the equation of motion:
y = y₀ + Vy₀t - (1/2)gt²
Since the initial vertical velocity (Vy₀) is zero (the ball is hit horizontally), the equation simplifies to:
y = y₀ - (1/2)gt²
Step 3: Solve for the acceleration due to gravity (g):
We know that g = 9.8 m/s² (approximate value on Earth).
Step 4: Substitute the known values into the equation of motion and solve for the maximum height (H):
y = y₀ - (1/2)gt²
Substituting y₀ = 1 m, g = 9.8 m/s², and t = 3 s, we get:
1 = 1 - (1/2)(9.8)(3)²
Simplifying further, we have:
1 = 1 - (1/2)(9.8)(9)
The equation becomes:
1 = 1 - 44.1
Solving for the maximum height (H), we find:
H = 45.1 meters above the ground
Therefore, the greatest height the baseball reached above the ground is approximately 45.1 meters.