A major leaguer hits a baseball so that it leaves the bat at a speed of 32.5 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 32.5 m/s and at an angle of 36.1 above the horizontal. You can ignore air resistance. (parts A-G please)

part A -At what two times is the baseball at a height of 10.7

m above the point at which it left the bat?

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

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

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

Part E-Calculate the vertical component of the baseball's velocity at a later time calculated in part (a)

Part F -What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?

Part G -What is the direction of the baseball's velocity when it returns to the level at which it left the bat?

Explanation / Answer

Here ,

a)

initial speed , u = 32.5 m/s

theta = 36.1 degree

let the time is t

using second equation of motion

h = u*t + 0.5 at^2

10.7 = 32.5 * sin(36.1) * t - 0.5 * 9.8 * t^2

solving for t

t = 0.68 s , 3.23 s

two times are 0.68 s and 3.23 s

b)
as the horizontal component of velocity of projectile is constant

at the earlier time

horizontal component of velocity = 32.5 * cos(36.1)

horizontal component of velocity = 26.3 m/s

the horizontal component of velocity at earlier time is 26.3 m/s

c)

Now , vertical component of velocity = u + a * t

vertical component of velocity = 32.5 * sin(36.1) - 9.8 * 0.68

vertical component of velocity = 12.5 m/s

the vertical component of velocity is 12.5 m/s

d)

as the horizontal component of velocity of projectile is constant

at the later time

horizontal component of velocity = 32.5 * cos(36.1)

horizontal component of velocity = 26.3 m/s

the horizontal component of velocity at the later time is 26.3 m/s

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