Question

A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.5m/s, starting from an initial position of 1.2m above the ground. When the watermelon reaches the peak of its flight, what is the velocity, the acceleration, the elapsed time, and the height above the ground?

Answer 1

AT the peak, its veloicty is zero, acceleration is -9.8m/s^2

vf^2=vi^2+2ad
d is the distance it traveled.
0=9.5^2-2*9.8*d
solve for d, then the height above ground is d+1.2m

Answer 2

time up = t = v / g

at peak
... v = 0 ... a = g

h = [vi^2 / (2 g)] + 1.2