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Arada [10]
3 years ago
11

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 obj

ect is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H?
Physics
1 answer:
Serhud [2]3 years ago
5 0

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 = \frac{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 = \frac{1}{2} g (t-1)^2

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

H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g     ................2

now subtract equation 2 from equation 1 so we get

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

53.2 = gt - \frac{1}{2} g

53.2 = 9.81 t - \frac{1}{2} 9.8

t = 5.92 s

so from equation 1

H = \frac{1}{2} (9.81)5.92^2

H = 171.90 m

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