A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 and

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 and at an angle of 34.9 above the horizontal. You can ignore air resistance.

A)At what two times is the baseball at a height of 8.00 above the point at which it left the bat?

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

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

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

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

Explanation / Answer

vo=30 m/s, =34.9o, y(t)=voy*t-1/2*g*t2. t1=0.554 t2=2.949

vou=vo*sin()17.16 m/s

A) t1=0.554 s, t2=2.949 s

B) vox=vo*cos()24.605 m/s is constant for the entire flight

C) vy=voy-g*t=17.16-9.8*0.55411.74 m/s, 17.16-9.8*2.949-11.74 m/s

D) 30 m/s

E) 0 velocity, it has stopped

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