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

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