Question #3047

A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.00 m/s, starting from an initial position 1.20 meters above the ground. When the watermelon reaches the peak of its flight, what is (a) its velocity, (b) its acceleration, (c) the elapsed time, and (d) its height above the ground?

Expert's answer

Let's use the equations of the motion under constant acceleration:

a) v=0

b) a=g=-9.8 m/s^{2}

c) t=v_{0}/g=9/9.8=0.92 s

d) H=h + v^{2}/ 2g =1.2+9^{2}/ (2* 9.8)=5.3 m

a) v=0

b) a=g=-9.8 m/s

c) t=v

d) H=h + v

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