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

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~