To find the time the rock remains in the air, we can use the kinematic equation:
h = h0 + v0t - (1/2)gt^2
Where:
h = final height (0 feet, as it reaches back to the ground)
h0 = initial height (6 feet)
v0 = initial velocity (55 feet/second, upward)
g = gravitational acceleration (32 feet/second^2, downward)
t = time
Setting h = 0, h0 = 6, v0 = 55, and g = 32, we have:
0 = 6 + 55t - (1/2)32t^2
Rearranging the equation:
16t^2 - 55t - 12 = 0
We can solve this quadratic equation to find the time the rock remains in the air. Using the quadratic formula:
t = (-b ยฑ โ(b^2 - 4ac)) / (2a)
Where:
a = 16
b = -55
c = -12
Calculating:
t = (-(-55) ยฑ โ((-55)^2 - 4 * 16 * -12)) / (2 * 16)
t = (55 ยฑ โ(3025 + 768)) / 32
t = (55 ยฑ โ(3793)) / 32
Therefore, the rock will remain in the air for approximately 3.03 seconds or 0.47 seconds (rounded to two decimal places).
To find the maximum height, we can use the formula for vertical displacement:
y = y0 + v0t + (1/2)gt^2
Since we are interested in the maximum height, we need to find the time at the peak of the trajectory. The time at the peak can be found using the formula:
v = v0 + gt
At the peak, the velocity (v) is zero, so we can rearrange the equation:
0 = 55 + (32)(t_peak)
Solving for t_peak:
t_peak = -55/32
Substituting this value back into the displacement formula:
y = y0 + v0(-55/32) + (1/2)(32)(-55/32)^2
y = 6 - (55)(55/32)(1/2)
y โ 407.03 feet (rounded to two decimal places)
The rock reaches a maximum height of approximately 407.03 feet after approximately 0.47 seconds (rounded to two decimal places).