A watermelon cannon fires a watermelon vertically up into the air at a velocity of +9.5m/s m/s, starting from an initial position 1.2 meters above the ground. When the watermelon reaches the peak...

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

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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

Hello!

The height of the watermelon above the ground is

`H(t)=H_0+V_0*t-(g*t^2)/2,`

where `H_0=1.2m` is the initial height, `V_0=9.5m/s` is the initial upward velocity and `g=9.8m/s^2` is the gravity acceleration.

Its velocity is

`V(t)=V_0-g*t`

and its acceleration is `g` (a constant).

The watermelon reaches the peak of its flight when H(t) has its maximum. It is reached at `t_1=V_0/g` (the property of a quadratic function).

(a) its velocity at that time is zero, which isn't a surprise. Actually, zero velocity is the equivalent condition for "the watermelon reaches the peak of its flight".

(b) its acceleration is g all the way, and also at a peak.

(c) time elapsed is `t_1=(V_0)/(g) = (9.5)/(9.8) approx` 0.97 (s).

(d) the height above the ground is `H(t_1)=1.2+9.5*0.97-(9.8/2)*(0.97)^2 approx` 5.8 (m).

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