A major leaguer hits a baseball so that it leaves the bat at a speed of 31.8 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 31.8 m/s and at an angle of 34.5 above the horizontal. You can ignore air resistance.

a) At what two times is the baseball at a height of 10.0 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?

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

Explanation / Answer

first the x - component is Vx = V cos34.5 = 31.8 X cos34.5 = 26.2 m/s

and y - component is Vy = V sin34.5 = 18 m/s

a )

h = Vyt - 1/2 g t2

0 = 1/2 g t2 - Vy t + h

0 = 4.9 t2 - 18 X t + 10

4.9 t2 - 18 t + 10 = 0

t = 2.99 and 0.682 sec

b )

Vx = V cos34.5

= 31.8 X cos34.5

= 26.2 m/s

c )

V = Vy - g t

V = 18 - 9.8 X 0.682

V = 11.31 m/sec

d )

Vx = V cos34.5

= 31.8 X cos34.5

= 26.2 m/s

e )

V = Vy - g t

V = 18 - 9.8 X 2.99

V = - 11.31 m/sec

f )

the magnitude of the baseball's velocity when it returns to the level at which it left the bat is same that is

31.8 m/sec

g )

the angle is

- 34.5

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