When the baseball is shot straight upward with an initial speed of 20 \rm m/s, what is the maximum height above its initial location?
H= v²/2g
20m
To find the maximum height above the initial location when a baseball is shot straight upward with an initial speed of 20 m/s, we can use the equations of motion.
First, we need to determine the time it takes for the baseball to reach its maximum height. We can use the equation:
vf = vi + at
Where:
- vf is the final velocity (at the maximum height)
- vi is the initial velocity
- a is the acceleration due to gravity (approximately -9.8 m/s^2, negative value indicates acceleration in the opposite direction of motion)
- t is the time
Since the ball is shot straight upward, its final velocity (vf) at the maximum height is 0 m/s. The initial velocity (vi) is 20 m/s, and the acceleration (a) is -9.8 m/s^2. By rearranging the equation, we can solve for the time (t):
0 = 20 - 9.8t
Solving for t:
9.8t = 20
t ≈ 2.04 seconds
Therefore, it takes approximately 2.04 seconds for the ball to reach its maximum height.
Next, we can use the equation for displacement:
Δy = vi * t + 0.5 * a * t^2
Where:
- Δy is the change in vertical position (displacement or height)
- vi is the initial velocity
- t is the time
- a is the acceleration due to gravity
Plugging in the values:
Δy = 20 * 2.04 + 0.5 * (-9.8) * (2.04)^2
Δy ≈ 20.4 - 20.11
Δy ≈ 0.29 meters
Therefore, the maximum height above its initial location for the baseball is approximately 0.29 meters.