Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 33.0 m/s and at an angle of 37.3 ∘ above the horizontal. You can ignore air resistance a.At what two times is the baseball at a height of 10.2 m above the point at which it left the bat? b. Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part (a). c. Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a). d.Calculate the horizontal component of the baseball's velocity at a later time calculated in part (a) e. Calculate the vertical component of the baseball's velocity at a later time calculated in part (a) f. What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?

Answer 1

Given launch velocity,

vertical launch velocity,

horizontal launch velocity,

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a.At what two times is the baseball at a height of 10.2 m above the point at which it left the bat?

Use formula

The ball is going up and g is acting downwards, so put negative sign.

Solve the quadratic equation using a calculator

ANSWER:

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b. Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part (a)

There is no acceleration in horizontal direction, so horizontal velocity remains constant

ANSWER :

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c. Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a)

Use formula

ANSWER :

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d.Calculate the horizontal component of the baseball's velocity at a later time calculated in part (a)

There is no acceleration in horizontal direction, so horizontal velocity remains constant

ANSWER :

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e. Calculate the vertical component of the baseball's velocity at a later time calculated in part (a)

Use formula

ANSWER :

Negative sign indicates direction is downwards

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f. What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?

First find time of flight

Use formula

Use formula

Negative sign indicates direction is downwards

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There is no acceleration in horizontal direction, so horizontal velocity remains constant

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ANSWER:

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