1 answer:
You might be interested in
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.
So hmm, if you notice the picture below
"y" would be the opposite to the 10° angle, and "x" would its adjacent
Reno is "y" north and "x" west from Miami
solve for "y", to see how much north
and
solve for "x" to see how much west
make sure, that in both cases, your calculator is in Degree mode, since the angle is in degrees, as opposed to Radian mode
now... for the section B)
if notice the Reno-Miami line, and you were to draw a North line through it, anywhere in between, or at an endpoint like the in the picture at Reno, the "clockwise" angle, will always be the same, 100°
bear in mind ( no pun intended ), that Bearings use the North line, and the clockwise angle from it, as in the picture below
"y" would be the opposite to the 10° angle, and "x" would its adjacent
Reno is "y" north and "x" west from Miami
and
make sure, that in both cases, your calculator is in Degree mode, since the angle is in degrees, as opposed to Radian mode
now... for the section B)
if notice the Reno-Miami line, and you were to draw a North line through it, anywhere in between, or at an endpoint like the in the picture at Reno, the "clockwise" angle, will always be the same, 100°
bear in mind ( no pun intended ), that Bearings use the North line, and the clockwise angle from it, as in the picture below