Question

A body is dropped from the top of a tower during the last second of its motion it travels 16 by 25 of the height of the tower calculate the height of the tower is equal to 10 metre per second square

Answer 1

Answer:

31.25 metres

Explanation:

Let height of the tower = H metres

Initial velocity (u) = 0 m/s. [bcz, the body is dropped]

According to equation,

h = ut + 1/2gt^2

where h is height, g is acceleration of gravity=10m/s^2, t is time taken.

H = 1/2×10×t^2

√(H/5) seconds = t........(i)

Let H' be the height covered in (t-1) seconds.

So, H' = H-(16/25)H

H' = 9H/25

9H/25 = 1/2×10×(t-1)^2

9H/125 = (t-1)^2

Square rooting both sides,

[3/5] × [√(H/5)] = t-1........(ii)

Put the value of (i) in (ii),

[3/5] × [√(H/5)] = √(H/5) -1

1 = [2/5] × [√(H/5)]

5/2 = √(H/5)

Squaring both sides,

25/4 = H/5

125/4 metres = H

Height of the tower = 31.25 metres

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Answer 2

Refers to attachment ♡♡~