1a) An object is thrown up with a velocity of 18m/s from a height of 520m.
What's the object's position after 3s?

Explanation:

An object is thrown up from a height of 520m

h_o = 520m

Initial velocity of thrown is 18m/s

u = 18m/s

What is the position of the object after 3seconds

t = 3s

Acceleration due to gravity

g = 9.81 m/s²

Let calculated the height the object will reached when thrown from the top.

Using equation of motion

h = ut + ½gt²

Since the body is thrown upward, it is acting against gravity then, gravity will be negative

Then,

h = ut — ½gt²

h = 18 × 3 — ½ × 9.81 × 3²

h = 54 — 44.145

h = 9.855 m

So, the body is above the top of the building at a distance of 9.855m

So, the total distance from the bottom is

Position = h + h_o

x = 9.855 + 520

x = 529.855m

x ≈ 530m,

The position of objects after 3seconds is 520m

6 cos 34, 6 sin 34 trig

it has to be b, about 97 percent.