Question

A stone is falling from top of tower covering distance 24.5 in last the second of its motion find night of the tower??? ​

Answer 1

Answer:

Explanation:let the stone is thrown from ground then the distance in first second is equal to dist. in last sec.

hope it help

Answer 2

Question :

A stone is dropped from top of tower covering a distance 24.5 m in last the second of it's motion. Find height of the tower?

Solution :

Here..

• Distance (s) = 24.5 m

• Initial velocity (u) = 0m/s

[As stone is falling freely]

• Acceleration due to gravity (a = g) = 10m/s²

________________ [ GIVEN ]

We know that ..

=> $S_{n}$ = u + $\dfrac{1}{2}$ a (2n - 1)

Put the known values in above formula

=> 24.5 = 0 + $\dfrac{1}{2}$ × 10 (2n - 1)

=> 24.5 = 5 (2n - 1)

=> 4.9 = 2n - 1

=> 4.9 + 1 = 2n

=> 5.9 = 2n

=> n = 2.95

=> n = 3 sec (approx.)

(Here n is time)

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Now..

=> s = ut + $\dfrac{1}{2}$ at²

=> s = (0)(3) +$\dfrac{1}{2}$ (10)(3)²

=> 5(9)

=> 45 m

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45 m is the height of the tower.

_____________ [ ANSWER ]

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》 If I take a = g = 9.8 m/s then

$S_{n}$ = u + $\dfrac{1}{2}$ a (2n - 1)

Put the known values in above formula

=> 24.5 = 0 + $\dfrac{1}{2}$ × 9.8 (2n - 1)

=> 24.5 = 4.9 (2n - 1)

=> 5 = 2n - 1

=> 5 + 1 = 2n

=> 6 = 2n

=> n = 3 sec

(Here n is time)

_______________________________

Now..

=> s = ut + $\dfrac{1}{2}$ at²

=> s = (0)(3) +$\dfrac{1}{2}$ (9.8)(3)²

=> 4.9(9)

=> 44.1 m

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44.1 m is the height of the tower.

_____________ [ ANSWER ]

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