Asked by Kathy
A watermelon cannon fires a watermelon vertically up into the air at a velocity of +11.5 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?
Answers
Answered by
drwls
(a) 0
(b) -9.8 m/s^2
(c) t = 11.5/9.8 = 1.1735 s
(d) Use the standard formula for position vs time, at time t
Y = 1.2 + 11.5 t - 4.9 t^2 (meters)
(b) -9.8 m/s^2
(c) t = 11.5/9.8 = 1.1735 s
(d) Use the standard formula for position vs time, at time t
Y = 1.2 + 11.5 t - 4.9 t^2 (meters)
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