Question

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?

Answer 1

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