A watermelon cannon fires a watermelon vertically up into the air at a velocity of + 9.50 m/s, starting from an initial position 1.20 meters above the ground. When the watermelon reaches the peak of its flight, what is the elapsed time in sec?

t = (V-Vo)/g = )-9.50)/-9.8 = 0.969 s.

To find the elapsed time when the watermelon reaches the peak of its flight, we can use the formulas of motion and the concept of symmetry in projectile motion.

The initial velocity of the watermelon, u = +9.50 m/s (positive because the watermelon is moving upwards).
The acceleration due to gravity, g = -9.8 m/s^2 (negative because it acts downwards).
The initial vertical displacement, s = +1.20 meters (positive because the watermelon starts 1.20 meters above the ground).

At the peak of its flight, the watermelon will momentarily stop moving vertically and start falling back down. This happens when its final vertical velocity, v = 0 m/s.

Using the formula for final velocity, v = u + gt,
0 = +9.50 + (-9.8)t.

We can rearrange this equation to solve for the time, t:

-9.8t = -9.50,
t = -9.50 ÷ -9.8,
t ≈ 0.969 sec.

Therefore, the elapsed time when the watermelon reaches the peak of its flight is approximately 0.969 seconds.

To find the elapsed time when the watermelon reaches the peak of its flight, we can use the kinematic equation for vertical motion:

v = u + at

where:
v = final velocity (at the peak, the velocity is 0 m/s)
u = initial velocity (9.50 m/s, positive because it is directed upwards)
a = acceleration (due to gravity, it is -9.8 m/s^2, negative because it acts downward)
t = time

Since the watermelon reaches its peak and then falls back down, we know the final velocity is 0 m/s. Plugging the values into the equation, we have:

0 = 9.50 m/s + (-9.8 m/s^2) * t

Simplifying the equation, we get:

9.8 m/s^2 * t = 9.50 m/s

Solving for t, we have:

t = 9.50 m/s / 9.8 m/s^2

t ≈ 0.9694 seconds

Therefore, when the watermelon reaches the peak of its flight, the elapsed time is approximately 0.9694 seconds.