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.
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.