[SOLVED] A packet is dropped from a stationary helicopter, hovering at a height 'h' from the ground level, reaches the ground in 12s. Calculate 1) value of 'h' "displacement" 2) final velocity of packet on reaching ground level take acceleration due to gravity=9.8ms^-2 Please show how you received your answer.

Use kinematic equations to solve:1) yf = yo + vo*t + 1/2at²yf = final heightyo = initial heightvo = initial velocitya = accelerationt = timeyf - yo = vo*t + 1/2at²yf - yo = hvo = 0Thus,h = 1/2at²h = 1/2(9.8)(12)² = 705.6 m2) vf = vo + atvo = 0Thus,vf = atvf = (9.8)(12) = 117.6 m/s

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