Draw 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= 2vₒ•sinα/g.
t(OA) =t/2 =2•v₀•sinα/2•g =v(oy)/g =>
v(oy) = g•t (OA)=9.8•3.2 = 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 = 31.36•0.5 – 9.8•0.5²/2 = 14.46 m.
(a) vAC: H-h=g•t(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) = 81•9.8/(31.36 +2.7•9.8) = 13.73 m/s.
X = 13.73•0.5= 6.86 m.
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.2 s after being hit. Then 2.7 s after reaching its maximum height, the ball barely clears a fence that is 81.0 m from where it was hit. Assume the ground is level. (a) What maximum height above ground level is reached by the ball? (b) How high is the fence? (c) How far beyond the fence does the ball strike the ground?
**So far I can only get part (a) which is 50.2 meters.
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