A baseball traveling upward passes a window 28.0 m above the street with a vertical speed of 13.0 m/s. If the ball was thrown from street level, what is the maximum height that the baseball reaches?
v = v i - g t
st top v = 0
so
t = vi/g to top
then
h = 28 + vi t - 4.9 t^2
To find the maximum height reached by the baseball, we can use the concept of kinematics and the fact that the vertical velocity of the ball becomes zero at the maximum height.
First, we need to determine the time it takes for the ball to reach the highest point. We can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (13.0 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)
t = time
Rearranging the equation, we get:
t = (v - u) / a
Substituting the given values, we have:
t = (0 m/s - 13.0 m/s) / (-9.8 m/s^2)
t ≈ 1.33 seconds
Now, we can find the displacement (change in height) of the ball at the maximum height using the equation of motion:
s = ut + (1/2)at^2
where:
s = displacement (change in height)
u = initial velocity (13.0 m/s)
t = time (1.33 seconds)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)
Substituting the given values, we have:
s = (13.0 m/s)(1.33 s) + (1/2)(-9.8 m/s^2)(1.33 s)^2
s ≈ 8.71 meters
Therefore, the maximum height reached by the baseball is approximately 8.71 meters.