Answer:5.29 seconds to hit ground 30.86 feet in air
Explanation:
The first equation, y = 7.16t^2 - 40t + 45, represents the vertical position of the cargo as a function of time.
The second equation, y = -0.047x^2 + 45, represents the height of the cargo as a function of horizontal distance traveled.
To find the time it takes for the cargo to hit the ground, we need to find when y = 0 in the first equation.
Setting y = 0, we have:
0 = 7.16t^2 - 40t + 45
Solving for t, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 7.16, b = -40, and c = 45.
t = (40 ± √(40^2 - 4 * 7.16 * 45)) / (2 * 7.16)
t = (40 ± √(1600 - 322.2)) / (2 * 7.16)
t = (40 ± √1277.8) / (2 * 7.16)
t = (40 ± 35.62) / 14.32
t = (75.62 / 14.32, 4.38 / 14.32)
Since t must be positive, t = (75.62 / 14.32) = 5.29 s
Now that we have the time it takes for the cargo to hit the ground, we can use that time in the second equation to find the horizontal distance the cargo traveled.
x = √((45 - 0) / -0.047)
x = √(45 / -0.047)
x = √(45 * -21.28) = √(957) = 30.86 ft
So, it takes the cargo 5.29 seconds to hit the ground, and it travels 30.86 feet horizontally.