You throw a baseball directly upward at time t = 0 at an initial speed of 12.3 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.

The maximum height the ball reaches above where it leaves your hand is 7.72 m

The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.

The given parameters;

initial velocity, u = 12.3 m/s

acceleration due to gravity, g = 9.8 m/s²

The maximum height the ball reaches above where it leaves your hand is calculated as;

The time for the ball to reach half of the maximum height is calculated as;

The time taken for the ball to pass half of its maximum height moving upwards is 2.14s and 0.37 s when moving downwards.

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Explanation:

At the maximum height, the ball's velocity is 0.

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)

x = 7.72 m

The ball reaches a maximum height of 7.72 m.

The times where the ball passes through half that height is:

x = x₀ + v₀ t + ½ at²

(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²

3.86 = 12.3 t - 4.9 t²

4.9 t² - 12.3 t + 3.86 = 0

Using quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8

t = 0.368, 2.14

The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.

not moving. pos accel,

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray