You throw a baseball directly upward at time t=0 at an initial speed of 13.1 m/s. What is the maximum height the ball reaches above where it leaves your hand? Ignore air resistance and take g=9.80 m/s2.
maximum height:
At what times does the ball pass through half the maximum height?
earlier time at half maximum height:
later time at half maximum height:
h = -4.90 t^2 + 13.1 t
t-max is on the axis of symmetry ... t = -b / 2a = -13.1 / (2 * -4.90)
... plug the time into the equation to find the max height
plug half the max height into the equation to find the times (up and down)
To find the maximum height reached by the ball, we can use the kinematic equation for vertical motion:
h = (v^2) / (2g)
Where:
h is the maximum height
v is the initial upward velocity (13.1 m/s)
g is the acceleration due to gravity (9.80 m/s^2)
Plugging in the values, we have:
h = (13.1^2) / (2 * 9.80)
h = 8.66 meters
Therefore, the maximum height the ball reaches above where it leaves your hand is approximately 8.66 meters.
Now, to determine the times at which the ball passes through half of the maximum height, we need to use the equation for vertical displacement:
h = v0t + (1/2)gt^2
Where:
h is the vertical displacement
v0 is the initial velocity (13.1 m/s)
g is the acceleration due to gravity (9.80 m/s^2)
t is the time
Since we are interested in the times when the ball is at half the maximum height, we can substitute h with (1/2) * 8.66 = 4.33 meters.
4.33 = (13.1)t + (1/2)(9.80)t^2
Rearranging the equation gives:
4.9t^2 + 13.1t - 4.33 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Using the values a = 4.9, b = 13.1, and c = -4.33, we can calculate the two times:
t = (-13.1 ± √(13.1^2 - 4 * 4.9 * -4.33)) / (2 * 4.9)
Simplifying the equation gives:
t ≈ -2.15 seconds or t ≈ 0.447 seconds
Since the time cannot be negative, the earlier time at half the maximum height is approximately 0.447 seconds, and the later time at half the maximum height is 2.15 seconds.