What is the final velocity and time until impact of an object dropped from 411.5 meters?

Question

using d=1/2AT^2

Answer 1

d = 1/2 At^2

411.5 = 1/2 (9.8 m/s^2) t^2

411.5 / {1/2 (9.8)} = t^2

t * 9.8 = Vfinal

Answer 2

First find the time required:

h = g*t^2/2, where g = 9.91 m/s^2 the acceleration of gravity, h the height ant t the time. Solve for t:

t = sqrt(2h/g), t = sqrt(2*411.5/9.81), t = 9.16 s

Then find the velocity:

v = g*t, v = 9.81*9.16, v = 89.85 m/s

Assuming that you neglect the resistance of the air and that you drop the object with zero initial velocity.

Answer 3

you can also use the equation:

Vf² = Vi² + 2ad

Vf = final velocity
Vi = initial velocity
a = acceleration
d = displacement

plug and chug

Vf² = 0 + 2(-9.8)(-411.5)
Vf = √(8065.4)
Vf = 89.8 m/s