A major leaguer hits a baseball so that it leaves the bat at a speed of 28.5 and

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.5 and at an angle of 36.9 above the horizontal. You can ignore air resistance. At what two times is the baseball at a height of 9.00 above the point at which it left the bat?

Explanation / Answer

s = ut+0.5at^2 s = 9 u = 28.5*sin(Radians(36.9)) = 17.112 a = 9.81 =>9 = 17.112t+0.5*9.81*t^2 =>4.905t^2+17.112t-9 = 0 =>t = 0.4642seconds.........First time at which height is 9m Maximum height time = u/9.81 = 1.744sec Maximum height = 14.925m 14.925-9 = 0.5*9.81*t^2 =>t = 1.099sec Second time at which height is 9m = (1.099+1.744) = 2.8434sec

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