Question

A certain freely falling object requires 1.1 s to travel the last 28 m before it hits the ground. (a) What was the velocity of the object when it was 28 m above the ground?(b) From what height above the ground did it fall?

Answer 1

Given,

Time, t = 1.1 sec

A)

Using equation of motion,

s = ut+ 0.5gt^2

Initial speed, u = - (s/t) + 0.5 gt

u = - (28/1.1) - 0.5 x 9.81 x 1.1

Initial speed, u = - 20.06 m/s

Magnitude is 20.06 m/s, negative sign shows direction.

B)

Height, h = (v^2/2g) + s

h = ((20.06)^2/ 2 x 9.81) + 28 = 48.51 m

Comment in case any doubt please rate my answer....