a.
From equations of motions under gravity
[tex]h = ut + \frac{1}{2} g {t}^{2} [/tex]
where h = height of building
u = initial velocity
t = time taken
g = acceleration due to gravity
[tex]h = 0 \times 1.9 + \frac{1}{2} \times 10 \times {1.9}^{2} \\ \\ h = 18.05m[/tex]
b.
[tex] {v}^{2} = {u}^{2} + 2gh[/tex]
v = final velocity which is what we need at the moment
[tex] {v}^{2} = {0}^{2} + 2 \times 10 \times 18.5 \\ \\ v = \sqrt{361} = 19m {s}^{ - 2} [/tex]
a. The height of the building is  18.05 m.
b. The rock was going to hit the ground with velocity is 19 m/s.
c. When assumed there is air resistance, the velocity is smaller than the calculated.
When an object is released from rest in free air considering the motion is depend only on the acceleration due to gravity, g.
You stand at the top of a tall building with a stopwatch. You drop a rock off the side of the building, and it takes the rock 1.9 seconds to hit the ground.
From second equation of motion while free falling
h = ut + 1/2 gt²
where h = height of building, u = initial velocity, t = time taken and g = acceleration due to gravity
Plug the values, we get
h = 0 x 1.9 + 1/2 x 9.81 x (1.9)²
h = 18.05 m
Thus, the height of the building is  18.05 m.
b. The velocity with which rock falling is
v = √2gh
where, v is the final velocity
Plug the values, we have
v = √(2x 9.81 x 18.05)v = 19 m/s.
Thus, the velocity when rock hits the ground is 19 m/s.
c. Assuming the air resistance, the velocity will be smaller than the calculated.
Learn more about free falling.
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