A bat makes contact with a ball 0.8 meters above the ground. The ball then travels with a speed of 22 m/s, moving up. How high does it go?
straight up I guess
(1/2) m v^2 = m g h
h = 22^2/(2*9.81)
total height = .8 + h
To determine how high the ball goes, we can use basic principles of projectile motion. We know the initial vertical velocity (upward) is given as 22 m/s, and the ball's starting height is 0.8 meters above the ground.
The first step is to calculate the time it takes for the ball to reach its highest point. At the highest point, the vertical velocity will be zero. We can use the following equation of motion:
v = u + at
Where:
v = final velocity (0 m/s at the topmost point)
u = initial velocity (22 m/s)
a = acceleration (in this case, the acceleration due to gravity, which is approximately -9.8 m/s^2)
t = time (unknown)
Rearranging the equation, we have:
t = (v - u) / a
Substituting in the given values:
t = (0 - 22) / -9.8
t = 22 / 9.8
t ≈ 2.24 seconds
So it takes approximately 2.24 seconds for the ball to reach its highest point.
Next, we can determine the maximum height achieved by the ball. We can use the following equation:
h = u * t + (1/2) * a * t^2
Where:
h = height at the topmost point (unknown)
u = initial velocity (22 m/s)
t = time (2.24 seconds)
a = acceleration (approximately -9.8 m/s^2)
Substituting the known values:
h = 22 * 2.24 + (1/2) * -9.8 * (2.24^2)
h ≈ 24.75 meters
Therefore, the ball reaches a height of approximately 24.75 meters above the ground.