You throw a baseball directly upward at time t=0 at an initial speed of 13.1 m/s

Question

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? At what times does the ball pass through half the maximum height? Ignore air resistance and take g = 9.80 m/s^2 Sapling Learn Learning macmillan learning You throw a baseball directly upward at time t s 0 at an initial speed of 13.1m/s. What is the maximum h height the ball reaches above where it leaves your hand? At what times does the ball pass through maximum height? Ignore air resistance and take g 9.80 m/s2 half the Maximum height: Number im Earlier time at half maximum height: Number Later time at half maximum height: Number Previous Give Up & View Solution O Check Answer 0 Next-1 Exit- Hint

Explanation / Answer

let the maximum height be h

by third equation of motion

v^2 = u^2 + 2as

0 = 13.1^2 - 2 * 9.8 * s

s = 8.7556 m

maximum height = 8.7556 m

by second equation of motion

s = ut + 0.5 * at^2

8.7556 / 2 = 13.1 * t - 0.5 * 9.8 * t^2

t = 0.39152 s

ball will reach half maximum height at t = 0.39152 s

by first equation of motion

v = u + at

0 = 13.1 - 9.8 * t

total time t = 1.3367

time took to cover other half = 1.3367 - 0.39152

time took to cover other half = 0.94518 s

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