
Herb K. answered 07/20/18
Tutor
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Semi-Retired College Professor - MIT Grad - very patient, experienced
h(t) = -4.9t^2 + 12.8t; h(t) = 0 = (-4.9t + 12.8)t; then: t = 0 (to start); and -4.9t + 12.8 = 0 (to finish);
then: t = 12.8/4.9 = 2.6122449 sec = 2.61 sec
note: V_0y = 12.8; also: [V(y)]^2 = [V_0y]^2 - 2(9.8)H; at top of flight path: V(y) = 0; then: 0 = 12.8^2 - 2(9.8)H
or: H = 12.8^2/19.6 = 8.36 meters