You throw a baseball directly upward at time t = 0 at an initial speed of 12.5 m/s. What is the maximum height the ball reaches above where it leaves your hand? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s2.
V^2 = Vo^2 + 2g*h
h = (V^2-Vo^2)/2g
hmax = (0-12.5^2)/-19.6 = 8 m.
h = Vo*t + 0.5g*t^2 = 8/2 = 4 m
12.5*t + (-4.9)t^2 = 4
-4.9t^2 + 12.5t - 4 = 0
t = 0.375, and 2.18 s.
(Use Quadratic Formula).
To determine the maximum height the ball reaches, we can use the kinematic equation for vertical motion:
Height (h) = Initial height + (Initial velocity * time) - (0.5 * acceleration * time^2)
Since the ball is thrown directly upward, the initial height is 0, the initial velocity is +12.5 m/s (positive because it is upward), and the acceleration is -9.80 m/s^2 (negative because it is in the opposite direction as the upward motion).
Let's calculate the maximum height:
Height = 0 + (12.5 * t) - (0.5 * 9.80 * t^2)
To find the maximum height, we need to find the time at which the height is maximized. At the maximum height, the velocity becomes 0. So, we can set the equation for velocity equal to 0:
Velocity = Initial velocity + (acceleration * time)
0 = 12.5 - 9.80 * t
Solving for t:
9.80 * t = 12.5
t = 12.5 / 9.80
t ≈ 1.28 s
Now we can substitute this time back into the height equation:
Height = 0 + (12.5 * 1.28) - (0.5 * 9.80 * (1.28^2))
Height ≈ 8.05 m
Therefore, the maximum height the ball reaches above where it leaves your hand is approximately 8.05 meters.
To determine the times at which the ball passes through half the maximum height, we need to find the time values when the height is equal to half the maximum height.
Height = 0.5 * maximum height
Let's substitute the maximum height we calculated earlier:
0.5 * 8.05 = 0 + (12.5 * t) - (0.5 * 9.80 * t^2)
This equation is a quadratic equation, and we can solve it using the quadratic formula. Let's rearrange the equation:
0.5 * 8.05 = 12.5 * t - 0.5 * 9.80 * t^2
-0.5 * 9.80 * t^2 + 12.5 * t - 0.5 * 8.05 = 0
Using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
a = -0.5 * 9.80, b = 12.5, c = -0.5 * 8.05
Substituting the values into the quadratic formula:
t = (-12.5 ± √(12.5^2 - 4 * (-0.5 * 9.80) * (-0.5 * 8.05))) / (2 * (-0.5 * 9.80))
Now we can solve for t to find the times when the ball passes through half the maximum height.