Answer:
(a) 7.5 seconds
(b) The horizontal distance the package travel during its descent is 1737.8 ft
Step-by-step explanation:
(a) The given function for the height of the object is s = -16·t² + v₀·t + s₀
The initial height of the object s₀ = 900 feet
The initial vertical velocity of the object v₀
= 0 m/s
The time it takes the package to strike the ground is found as follows;
0 = -16·t² + 0×t + 900
900 = 16·t²
t² = 900/16 = 62.25
t = √62.25 = 7.5 seconds
(b) Given that the horizontal velocity of the package is given as 158 miles/hour, we have
158 miles/hour = 231.7126 ft/s
The horizontal distance the package covers in the 7.5 second of vertical flight = 231.7126 ft/s × 7.5 s = 1737.8445 feet = 1737.8 ft to one decimal place
The horizontal distance the package travel during its descent = 1737.8 ft.
5s + 5l = 155, 10s + 12l = 346
Solve for s in the first equation.
5s = 155 - 5l
s = 31 - l
Substitute s into the second equation.
10(31 - l) + 12l = 346
310 - 10l + 12l = 346
310 + 2l = 346
2l = 36
l = 18
Substitute l into the first equation.
5s + 5(18) = 155
5s + 90 = 155
5s = 65
s = 13
s = 13, l = 18
4x+6=3x+10 so x=4, to find the measure of KJL, plug in x into the equation 4x+6. angle KJL=22°
26*37=962sqft
Add 4 ft to both numbers (2ft per side)
30*41=1230sqft
Subtract out the garden area
1230-962=268sqft
The path is 268 sq ft