A batter hits a baseball m = 0.41 kg from rest with a force of F = 81

N, The ball is in contact with the bat for t = 0.22 s. It is observed that the ball leaves the bat

with an angle of 0 = 29 degrees with respect to the horizontal.

How long was the ball in the air if it was hit from the top of the cliff with a height of h=86m?

7.0s

Explanation:

Mass = 0.41kg

F= 81N

t = 0.22s

¤ = 29°

Lo = 86m

From impulse equation,

F*t = m* v

81 * 0.22 = 0.41 * v

Vo = 17.82 / 0.41

Vo = 43.46m/s

Vx= velocity across horizontal plane

Vy = velocity across vertical plane

Vx = Vo * cos ¤

Vy = Vo * sin ¤

Vx = 43.46 * cos 30° = 37.64 m/s

Vy = 43.46 sin 30° = 21.73 m/s

Distance travelled across the vertical plane,

L = Lo + Vy *t + ½gt²

0 = 86 + 21.73t - 4.9t²

4.9t² - 21.73t - 86 = 0

Solving for t in the quadratic equation,

t = 6.96 or -10.04

Using the positive root since time can't be negative, t = 6.96 approximately 7.0s

c)as speed changes, wavelength and frequency change.

everything changes!

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