A major leaguer hits a baseball so that at a speed of 30.5 m/a and at an angle o

Question

A major leaguer hits a baseball so that at a speed of 30.5 m/a and at an angle of 37 3 degree above the horizontal. You can ignore air resistance. At what two times is the baseball at a height of 11.2 m above the point at which it left the bat? Give your answers is ascending order separated with comma. t_1, t_3 = Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part(a). v_z = 24.3 m/s Calculate the vertical component of the baseball's velocity at an earlier time calculated in part (a) v_p = m/s

Explanation / Answer

Y = Uyt - 0.5gt^2

-11.2 = 30.5 sin 37.3*t + 0.5*9.8*t^2

-11.2 = 18.48t + 4.9t^2

4.9t^2 + 18.48t + 11.2 = 0

t = 18.48 +- sqrt(18.48^2 - 4*4.9*11.2) / 9.8

t1 = 3.01s

t2 = 0.7586 s

Vx = Ux (as there is no acceleration in x- direction )

Vx at t1 and t2 = 30.5 cos 37.3 = 28.1 m/s

Vy = Uy - gt1 = (30.5sin 37.3 - 9.8*0.7586) = 11.04 m/s

Vy = Uy + gt = (30.5sin 37.3 + 9.8*3.01) = 48 m/s

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