Answer:
x = 727.5 km
Explanation:
With the conditions given using trigonometry, we can find the tangent
tan θ = CO / CA
With CO the opposite leg and CE is the adjacent leg which is the distance from the Tierral to Sun
D =150 10⁶ km (1000m / 1 km)
D = 150 10⁹ m.
We must take the given angle to radians.
1º = 3600 arc s
π rad = 180º
θ = 1 arc s (1º / 3600 s arc) (pi rad / 180º) =
θ = 4.85 10⁻⁶ rad
That angle is extremely small, so we can approximate the tangent to the angle
θ = x / D
x = θ D
x = 4.85 10-6 150 109
x = 727.5 103 m
x = 727.5 km
These are the correct solutions:
It is 11 a.m. in the Eastern Time Zone; therefore, it is 8 a.m. in the Pacific Time Zone. (3 hrs behind)
It is 3 p.m. in the Central Time Zone; therefore, 2 p.m. in the Mountain Time Zone. (1 hr behind)
It is 6 p.m. in the Pacific Time Zone; therefore, it is 4 p.m in Hawaii. (2 or 3 hours behind depending on time of year)
It is 6 p.m. in Hawaii; therefore, it is 11 p.m. in the Eastern Time Zone (5 or 6 hours behind depending on time of year).
It is 3 p.m. in Hawaii; therefore, it is 6 p.m. in the Mountain Time Zone (3 or 4 hours behind depending on time of year).
So....
imagine the x and y axis or draw it on a piece of paper, a white board or even on your wall if you like. I'm kidding just draw it on a piece of paper
50m up and 20m down
the dog's displacement is 50 - 20 which is 30
your answer is 30m north
Initial speed of Gazelle is along x direction and its value will be

also its initial height is given as

Part a)
now from kinematics along Y direction

as we know that





Part b)
distance moved horizontally

as we know that

now we will have

so it will lend at distance of 4 m.
Part c)
final velocity in vertical direction



so net speed will be


