Question

a major leaguer hits a baseball so that it leaves the bat at a speed of 28.5m/s at an angle of 35.9 degrees above the horizontal. you can ignore air resistance. A) at what two times in the baseball at a

height of 10.6m above the point at which it left the bat? express your answer in seconds.

B) calculate the horizontal component of the baseball’s velocity at an earlier time calculated in A. express your answer in meter per second.

C) calculate the vertical component of the baseball’s velocity at an earlier time calculated in A. express your answer in meter per second.

D) calculate the horizontal component of the baseball’s velocity at a later time calculated in A. express your answer in meter per second.

E) calculate the vertical component of the baseball’s velocity at a later time calculated in A. express your answer in meter per second.

F) what is the magnitude of the baseball’s velocity when it returns to the level at which ti left the bat? express your answer in meter per second.

G) what is the direction of the baseball’s velocity when it returns to the level at which it left the bat? express your answer in degrees.

Answer 1

Given launch velocity,

Vertical launch velocity,

Horizontal launch velocity,

A) At what two times in the baseball at a height of 10.6 m above the point at which it left the bat?

Consider the vertical motion of the ball

Use formula

Solve the quadratic equation using a calculator

ANSWER : and

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B) Calculate the horizontal component of the baseball’s velocity at an earlier time calculated in A.

Consider the horizontal motion of the ball.

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 A.

Consider vertical motion of the ball

Use formula

ANSWER :

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D) Calculate the horizontal component of the baseball’s velocity at a later time calculated in A.

Consider the horizontal motion of the ball.

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 A.

Consider vertical motion of the ball

Use formula

At max height vertical velocity becomes zero.

ANSWER :

Negative sign indicates downwards direction

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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?

Consider the horizontal motion of the ball.

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

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Consider vertical motion of the ball

Consider rising part

Use formula

Consider falling part

Use formula

downwards

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Magnitude,

ANSWER :

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G) what is the direction of the baseball’s velocity when it returns to the level at which it left the bat?

Direction

ANSWER :

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