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?
To answer these questions, we need to analyze the motion of the watermelon. We can use the equations of motion from classical mechanics.
(a) To find the velocity of the watermelon at the peak of its flight, we need to note that the velocity at the peak is zero. This occurs when the watermelon reaches its maximum height and momentarily stops before falling back down.
(b) The acceleration experienced by the watermelon is due to the force of gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2, directed downwards.
(c) To find the elapsed time, we can use the fact that the time it takes to reach the peak is equal to the time it takes to fall back to the starting height. This is because the watermelon follows a symmetrical parabolic trajectory. Therefore, we can calculate the time it takes for the watermelon to reach the peak and then double that value.
(d) To find the height of the watermelon above the ground at the peak, we need to calculate the maximum height reached by the watermelon.
Now let's calculate the values:
(a) The velocity at the peak is zero, since the watermelon momentarily stops before falling back down.
(b) The acceleration experienced is due to gravity and is approximately 9.8 m/s^2, directed downwards.
(c) To find the elapsed time, we need to calculate the time it takes for the watermelon to reach its peak. We can use the equation:
v = u + at
where, v is the final velocity (zero in this case), u is the initial velocity (11.5 m/s), a is the acceleration (-9.8 m/s^2), and t is the time.
Rearranging the equation, we get:
t = (v - u) / a
Substituting the values, we get:
t = (0 - 11.5) / -9.8
t ≈ 1.17 seconds
Since the watermelon takes the same amount of time to reach the peak and fall back down, the elapsed time is 2 * 1.17 seconds ≈ 2.34 seconds.
(d) To find the height of the watermelon above the ground at the peak, we can use the equation for displacement:
s = ut + (1/2)at^2
where, s is the displacement, u is the initial velocity (11.5 m/s), a is the acceleration (-9.8 m/s^2), and t is the time.
Substituting the known values, we get:
s = (11.5 * 1.17) + (0.5 * -9.8 * 1.17^2)
s ≈ 6.8 meters
Therefore, at the peak of its flight, the watermelon has (a) zero velocity, (b) an acceleration of -9.8 m/s^2, (c) an elapsed time of approximately 2.34 seconds, and (d) a height above the ground of approximately 6.8 meters.
(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)