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?

Expert's answer

According to the equations of the motion

V=V_{0}-gt;

H = H_{0} + V_{0}t - gt^{2}/2

At the peak of the flight V = 0, so

V_{0} = gt

t = V_{0}/g

Thus

H = H_{0} + V_{0}^{2}/g - V_{0}^{2}/2g = H_{0} + V_{0}^{2}/2g = 1.2 + 11^{2}/(2*10) = 7.25 m

V=V

H = H

At the peak of the flight V = 0, so

V

t = V

Thus

H = H

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