Question

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).

Related Questions

To play baseball, you throw the ball. The expression 3m where m is the number of minutes played can... A(n) 0.163 kg baseball is thrown with a speed of 9.37 m/s. It is hit straight back at the pitcher... You throw a baseball straight up at 40 m/s and it reaches its peak after 4.00 seconds. What is its s... If I throw a baseball horizontally off of a 100m tower at a velocity of 3.33m/s how long will it tak... When baseball players throw the ball in from the outfield, they sometimes allow it to take one boun... A baseball pitcher can throw a base ball at 150 ft/s. Can this pitcher hit the 200-ft ceiling of its... You throw a baseball at a 45∘ angle to the horizontal, aiming at a friend who's sitting in a tree a... You throw a 238 g baseball straight up from a height of 1.73 m and it reaches a height of 13.2 m. Us... A baseball is throw from the top of a building and falls to the ground below. Its path is approximat... You throw a baseball directly upward at time t=0 at an initial speed of 13.1 m/s. What is the maximu...