A major leaguer hits a baseball so that it leaves the bat at a speed of 30.9 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.9 m/s and at an angle of 36.8 above the horizontal. You can ignore air resistance.
Part A At what two times is the baseball at a height of 11.4 m above the point at which it left the bat?
Part B Calclulate the horizontal component of the baseball's velocity at an earlier time calculated in part a?    Part C Calculate the vertical component of the baseballs velocity st an earlier time calculated in part a
Part D Calculate the horizontal component of the baseballs velocity at a LATER time calculated in part a
Part E Calculate the vertical component of the baseballs velocity at a LATER time calculated in part a
Part F What is the magnitude of the baseballs velocity when it returns to the level at which it left the bat?
Part G What is the direction of the baseballs velocity when it returns to the level at which it left the bat? Below the horizontal

Thank you very much A major leaguer hits a baseball so that it leaves the bat at a speed of 30.9 m/s and at an angle of 36.8 above the horizontal. You can ignore air resistance.
Part A At what two times is the baseball at a height of 11.4 m above the point at which it left the bat?
Part B Calclulate the horizontal component of the baseball's velocity at an earlier time calculated in part a?    Part C Calculate the vertical component of the baseballs velocity st an earlier time calculated in part a
Part D Calculate the horizontal component of the baseballs velocity at a LATER time calculated in part a
Part E Calculate the vertical component of the baseballs velocity at a LATER time calculated in part a
Part F What is the magnitude of the baseballs velocity when it returns to the level at which it left the bat?
Part G What is the direction of the baseballs velocity when it returns to the level at which it left the bat? Below the horizontal

Thank you very much
Part A At what two times is the baseball at a height of 11.4 m above the point at which it left the bat?
Part B Calclulate the horizontal component of the baseball's velocity at an earlier time calculated in part a?    Part C Calculate the vertical component of the baseballs velocity st an earlier time calculated in part a
Part D Calculate the horizontal component of the baseballs velocity at a LATER time calculated in part a
Part E Calculate the vertical component of the baseballs velocity at a LATER time calculated in part a
Part F What is the magnitude of the baseballs velocity when it returns to the level at which it left the bat?
Part G What is the direction of the baseballs velocity when it returns to the level at which it left the bat? Below the horizontal

Thank you very much

Explanation / Answer

u = 30.9 m/s, theta = 36.8 deg

a) horizontal component = u cos theta = 30.9 * cos 36.8 = 24.74 m/s

vertical component = u sin theta = 30.9 * sin 36.8 = 18.51 m/s

y = uy t - 1/2 g t2

11.4 = 18.51 t - 1/2 * 9.8 * t2

4.9 t2 - 18.51 t + 11.4 = 0

t = 0.774 s, t = 3.00 s

b) horizontal component = u cos theta = 30.9 * cos 36.8

= 24.74 m/s

c) vertical component vy = uy - g t = 18.51 - (9.8 * 0.774)

= 10.9 m/s

d) horizontal component = u cos theta = 30.9 * cos 36.8

= 24.74 m/s

e) vertical component vy = uy - g t = 18.51 - (9.8 * 3.00)

= -10.89 m/s

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