Part A At what two times is the baseball at a height of 10.2 m above the point a

Question

Part A

At what two times is the baseball at a height of 10.2 m above the point at which it left the bat?

Give your answers in ascending order separated with comma.

0.795,2.62

t1,t2 =

0.795,2.62

  s   Constants Part D A major leaguer hits a baseball so that it leaves the bat at a speed of 28.6 m/s and at an angle of 35.8, above the horizontal. You can ignore air resistance. Calculate the horizontal component of the baseball's velocity at the later of the two times calculated in part (a). m/s Submit Request Answer

Explanation / Answer

The vertical componet velocity of the ball is

uy = u sin theta = 28.6 m/s sin 35.8 = 16.72 m/s

Apply kinematic equation

y = uyt - 1/2 ay t^2

10.2 m = 16.72 t - 1/2 *( 9.8) t^2

4.9 t^2 -16.72 t + 10.2 m = 0

solving quadratic equatin

t 1, t2= 0.795s , 2.62 s

(D)

horizontal compoenent velcoity does not change with time . in entire motion of hte ball it remains constant

ux = u cos theta = 28.6 cos 35.8 = 23.19 m/s

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