If a freefall object requires 2s to travel the last 37.5m before hitting ground, What is its initial height?

Question

Please show steps! (and formula used!)

Answer 1

Let:

h be the initial height in metres,

t be the time in sec. to fall distance h metres,

g be the acceleration due to gravity in m/sec^2.

h = gt^2 / 2 ...(1)

For the first part of the fall:

h - 37.5 = g(t - 2)^2 / 2 ...(2)

From (2):

2(h - 37.5) = g(t - 2)^2

Substituting for h from (1):

2(gt^2 / 2 - 37.5) = g(t - 2)^2

gt^2 - 75 = g(t^2 - 4t + 4)

- 75 = - 4gt + 4g

t = (4g + 75) / (4g)

= 1 + 75 / (4g)

= 2.9113 sec.

Substituting this in (1):

h = 9.81 * 2.9113^2 / 2

= 41.6 m.