Answer to Question #1416 in Mechanics | Relativity for Krystyna

Question #1416
A watermelon cannon fires a watermelon vertically up into the air at a velocity of +11.0 m/s, starting from an initial position 1.20 meters above the ground. When the watermelon reaches the peak of its flight, what is its height above the ground?
1
Expert's answer
2011-01-28T17:32:59-0500
According to the equations of the motion
V=V0-gt;
H = H0 + V0t - gt2/2
At the peak of the flight V = 0, so

V0 = gt
t = V0/g
Thus
H = H0 + V02/g - V02/2g = H0 + V02/2g = 1.2 + 112/(2*10) = 7.25 m

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