Question

An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H? An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H?

Answer 1

H = 171.90 m

Step-by-step explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H =
`(1)/(2) gt^2` ......................1

here t is time and a is acceleration

so

we find t first

we know during time (t -1) s , it fall distance (H - 53.2) m

so equation of distance

`H - 53.2 = (1)/(2) g (t-1)^2`

`H - 53.2 = (1)/(2) g (t^2-2t+1)`

`H - 53.2 = (1)/(2) gt^2-gt+(1)/(2) g` ................2

now subtract equation 2 from equation 1 so we get

`H - (H - 53.2) =(1)/(2) gt^2- ((1)/(2) gt^2-gt+(1)/(2) g)`

53.2 = gt -
`(1)/(2) g`

53.2 = 9.81 t -
`(1)/(2) 9.8`

t = 5.92 s

so from equation 1

H =
`(1)/(2) (9.81)5.92^2`

H = 171.90 m