A major leaguer hits a baseball so that it leaves the bat at a speed of 29.4 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 29.4 m/s and at an angle of 37.1 above the horizontal. You can ignore air resistance.

1.) At what two times is the baseball at a height of 11.3 m above the point at which it left the bat?

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

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

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

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

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

7.) What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

(A) IN vertical,

v0 = 29.4 sin37.1 = 17.73 m/s

y = 11.3 m

a = - g = - 9.81 m/s^2

y = v0 t + a t^2 / 2

11.3 = 17.73t - 9.81 t^2 / 2

4.9 t^2 - 17.73t + 11.3 = 0

t = 0.83 sec And 2.79 sec

(2) horizontal conponent of velocity will always remain constant.

vx = 29.4 cos37.1 = 23.45 m/s

(3) vy = v0y + a t

v = 17.73 - (9.81 x 0.83) = 9.60 m/s

(4) vx = 23.45 m/s

(5) v = 17.73 - (9.81 x 2.79) = 9.6 m/s

(6) magnitude of velocity at same level will be same.

v = 29.4 m/s

(7) direction = 37.1 deg below the horizontal.

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