A major leaguer hits a baseball so that it leaves the bat at a speed of 30 m/s at an angle of 36.9° above the horizontal. The bat is at a height of 1.2 m when it hits the baseball. Ignore air resistance. a) What is the time of flight of the baseball? b) What is the maximum height of the baseball? c) What is the horizontal distance the baseball flies? d) What are the components of the velocity of the baseball just before it lands on the ground? e) What is the final magnitude and direction of the baseball? f) At what two times is the baseball at a height of 10 m? g) What are the components of the velocity at the two times in part (f) ?

Question

Maxim Miles

Physics

19 August, 18:45

A major leaguer hits a baseball so that it leaves the bat at a speed of 30 m/s at an angle of 36.9° above the horizontal. The bat is at a height of 1.2 m when it hits the baseball. Ignore air resistance. a) What is the time of flight of the baseball? b) What is the maximum height of the baseball? c) What is the horizontal distance the baseball flies? d) What are the components of the velocity of the baseball just before it lands on the ground? e) What is the final magnitude and direction of the baseball? f) At what two times is the baseball at a height of 10 m? g) What are the components of the velocity at the two times in part (f) ?

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Alexis · Answer 1

a) Initial velocity (along y axis) = 30 m/s

Vertical displacement = 1.2 m downwards

s = ut + 1/2 gt²

1.2 = - 30 sin 36.9 t +.5 x 9.8 t² (u is negative as it is in upward direction)

1.2 = - 18 t + 5 t²

5 t² - 18 t - 1.2 = 0

t = 3.67 s

b) v² = u² - 2gH (H is maximum height)

0 = (30 sin 36.9) ² - 2 x 9.8 x H

H = 16.5 m

c) horizontal distance will be covered with velocity 30 cos 36.9 m/s

time of flight is 3.67

Horizontal distance covered

= 3.67 x 30 cos 36.9

= 88 m

d) horizontal component will be same that is 30 cos 36.9 m/s

= 24 m / s

vertical component will be calculated as follows.

v = u + gt

v = - 30 sin 36.9 + 9.8 x 3.67

= - 18 + 36

= 18 m / s