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January 25, 2015

January 25, 2015

Posted by **Devin** on Tuesday, October 9, 2012 at 6:54pm.

**So far I can only get part (a) which is 50.2 meters.

- Physics -
**Elena**, Wednesday, October 10, 2012 at 3:14pmDraw the trajectory of the ball and and indicate the following points: starting point O; the max height A; the terminal point of trajectory B (where the ball hits the ground); C is the position where the ball is after 2.7 sec after the max position (the height of the fence); and point D which is located symmetrically to point C but on OA.

The time of motion from the initial (O) point to the terminal (b) point is t= 2vₒsinα/g.

t(OA) =t/2 =2v₀sinα/2g =v(oy)/g =>

v(oy) = gt (OA)=9.83.2 = 31.36 m/s.

t(DA) = t(AC) = 2.7 s. => t(OD) =t(OA)-t(DA) =3.2-2.7= 0.5 s.

The height of the point D= the height of the point C= h

(b) h = v(oy) t(OD) gt(OD)²/2 = 31.360.5 9.80.5²/2 = 14.46 m.

(a) vAC: H-h=gt(AC)²/2 = 35.72 m.

Max height H= h+35.72 = 14.46+32.72 = 50.18 m.

(c)

81 = L/2 + x = v(ox) 31.36/g +v(ox) 2.7

v(ox) (31.36/9.8 +2.7) = 81

v(ox) = 819.8/(31.36 +2.79.8) = 13.73 m/s.

X = 13.730.5= 6.86 m.

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