Question #15454

A watermelon cannon fires a watermelon vertically up into the air at a velocity of +12.0 m/s, starting from an initial position 1.20 meters above the ground. When the watermelon reaches the peak of its flight, what is (a) its velocity, (b) its acceleration, (c) the elapsed time, and (d) its height above the ground?

Expert's answer

(a) its velocity

The velocity of a watermelon has zero value at the peak of its flight.

(b) its acceleration

The acceleration of a watermelon has zero value at the peak of its flight too.

(c) the elapsed time

We can write down the next

V = V0 - gT,

where V is the current velocity of a watermelon, V0 = 12 m/s is the initial velocity of a watermelon, g is the gravity acceleration and T is the time elapsed after firing. As the velocity of a watermelon has zero value at the peak of its flight, we can find the time elapsed by watermelon to reach the peak:

V0 - gT = 0 ==> T = V0/g = 12[m/s]/9.8[m/s²] ≈ 1.22 s.

(d) its height above the ground?

H = H0 + gT²/2 = 1.20[m] + 9.8[m/s²]*(1.22[s])²/2 ≈ 8.49 m.

The velocity of a watermelon has zero value at the peak of its flight.

(b) its acceleration

The acceleration of a watermelon has zero value at the peak of its flight too.

(c) the elapsed time

We can write down the next

V = V0 - gT,

where V is the current velocity of a watermelon, V0 = 12 m/s is the initial velocity of a watermelon, g is the gravity acceleration and T is the time elapsed after firing. As the velocity of a watermelon has zero value at the peak of its flight, we can find the time elapsed by watermelon to reach the peak:

V0 - gT = 0 ==> T = V0/g = 12[m/s]/9.8[m/s²] ≈ 1.22 s.

(d) its height above the ground?

H = H0 + gT²/2 = 1.20[m] + 9.8[m/s²]*(1.22[s])²/2 ≈ 8.49 m.

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