Question

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.

Answer 1

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.