A major leaguer hits a baseball so that it leaves the bat at a speed of 32.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 32.0 m/s and at an angle of 36.5 above the horizontal. You can ignore air resistance.

Part A) At what two times is the baseball at a height of 9.00 m above the point at which it left the bat?

Part B) Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A.

Part C) Calculate the vertical component of the baseball's velocity at each of the two times you found in part A.

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

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

Explanation / Answer

Here ,

initial velocity , u = 32 m/s

theta = 36.5 degree

A)

let the time is t

Using second equation of motion

y = u*t + 0.5 at^2

9 = 32 * sin(36.5) * t - 0.5 * 9.8 * t^2

solving for t

t = 0.55 s , 3.33 s

the two times are 0.55 s , 3.33 s

B)

as the horiontal compoenent will be same at all times

vx = v * cos(36.5)

vx = 32 * cos(36.5)

vx = 25.7 m/s

the horiontal compoenent will be 25.7 m/s

C)

Now , magnitude of vertical velocity will be same

vy = 32 * sin(36.5) * - 9.8 * 0.55

vy = 13.6 m/s

the velocity at 0.55 is 13.6 m/s upwards

the velocity at 3.33 is 13.6 m/s downwards

D)

the magnitude of velocity is same as the initial speed

magnitude of velocity is 32 m/s

E)

direction of ball will be 36,5 m/s below horizontal

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.