Final answer:
The maximum height reached by the ball is 9.30 meters. The times it passes through half the maximum height on the way up and down (t1/2,up and t1/2,down) can be found using kinematic equations for uniformly accelerated motion with the provided initial speed and gravity.
Step-by-step explanation:
To determine the maximum height hmax reached by the baseball thrown directly upward and the times t1/2,up and t1/2,down it passes through half the maximum height on the way up and down, we can use the kinematic equations for uniformly accelerated motion. The equation we use for maximum height is h = v02 / (2g), where v0 is the initial velocity and g is the acceleration due to gravity.
Substituting 13.5 m/s for v0 and 9.80 m/s2 for g, we get:
hmax = (13.5 m/s)2 / (2 × 9.80 m/s2)
hmax = (182.25 m2/s2)/(19.6 m/s2)
hmax = 9.30 m
The time to reach the maximum height is found using t = v0 / g, while the times the ball passes through half the maximum height can be found by solving for the time when the height is hmax/2, using the kinematic equation h = v0t - 1/2gt2.
To find the maximum height reached by the ball, we can use the equation h = (v0^2) / (2g), where h is the maximum height, v0 is the initial velocity, and g is the acceleration due to gravity. Plugging in the given values, we have h = (13.5^2) / (2 * 9.80) = 9.09 m.
To find the times when the ball passes through half the maximum height on the way up and down, we can use the equation h = (v0 * t) - (1/2 * g * t^2). Rearranging this equation, we get t = [(v0/g) + sqrt((v0^2/g^2) + 2h/g)] / 2.
For the time on the way up, we have t₁₂,up = [(13.5/9.80) + sqrt((13.5^2/9.80^2) + 2*9.09/9.80)] / 2 ≈ 0.70 s.
For the time on the way down, we have t₁₂,down = [(13.5/9.80) - sqrt((13.5^2/9.80^2) + 2*9.09/9.80)] / 2 ≈ 1.32 s.