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November 28, 2014

November 28, 2014

Posted by **Devin** on Tuesday, October 9, 2012 at 6:26pm.

- Physics -
**Elena**, Wednesday, October 10, 2012 at 3:08pmDraw the trajectory of the ball 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)= 3.2s=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 = the height of the fence:

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

(a) 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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