Abaseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. a) how far away does the ball strike the ground? b) what is the maximum height of the ball?

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

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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.

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