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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?

Respuesta :

Answer:

H = 171.90 m

Explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H = [tex]\frac{1}{2} gt^2[/tex]    ......................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

[tex]H - 53.2 = \frac{1}{2} g (t-1)^2[/tex]

[tex]H - 53.2 = \frac{1}{2} g (t^2-2t+1)[/tex]

[tex]H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g[/tex]     ................2

now subtract equation 2 from equation 1 so we get

[tex]H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)[/tex]

53.2 = gt - [tex]\frac{1}{2} g [/tex]

53.2 = 9.81 t - [tex]\frac{1}{2} 9.8 [/tex]

t = 5.92 s

so from equation 1

H = [tex]\frac{1}{2} (9.81)5.92^2[/tex]

H = 171.90 m