Question

23. A stone is dropped from the top of a tower 400 m
high. At the same time, another stone is projected
vertically upward with a velocity 100 m s-1. Find
when and where the two meet.
[Ans. 78.4 m below top, 4 s from start]​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Height of Tower (H) = 400m.
• Speed of projection of the second ball = 100 m/s .
• Initial velocity of the First stone = 0 m/s.

$\huge{\bold{\underline{Explanation:-}}}$

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Let the distance travelled by the First Stone(stone A) be "x" and Second stone(stone B) be "y" before they meet.

#refer the attachment for figure.

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Now, Applying second kinematic equation for the First Stone,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\large{\tt \leadsto x = 0 \times t + \dfrac{1}{2}gt^2}$

$\large{\tt \leadsto x = 0 + \dfrac{1}{2}gt^2}$

$\large{\tt \leadsto x = \dfrac{1}{2}gt^2 \: ------(1)}$

$\large{\boxed{\tt x = \dfrac{1}{2}gt^2}}$

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Now, Applying Second kinematic equation for the Second Stone,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\large{\tt \leadsto y = 100 \times t + \dfrac{1}{2}(-g)t^2}$

$\large{\tt \leadsto y = 100t - \dfrac{1}{2}gt^2\: ------(2)}$

$\large{\boxed{\tt x = 100t - \dfrac{1}{2}gt^2}}$

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As we know,

x + y = 400 meters = Height of the Tower.

Now, Substituting the values,

$\large{\boxed{\tt x + y = 400}}$

$\large{\tt \leadsto \dfrac{1}{2}gt^2 + 100t - \dfrac{1}{2}gt^2 = 400}$

$\large{\tt \leadsto \cancel{\dfrac{1}{2}gt^2} + 100t - \cancel{\dfrac{1}{2}gt^2} = 400}$

$\large{\tt \leadsto 100t = 400}$

$\large{\tt \leadsto t = \dfrac{400}{100}}$

$\large{\tt \leadsto t = \dfrac{\cancel{400}}{\cancel{100}}}$

$\huge{\boxed{\boxed{\tt t = 4 \: seconds}}}$

So, the meet each other at 4 seconds.

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To calculate at what distance they meet,

Substituting the time value in equation (1),

$\large{\boxed{\tt x = \dfrac{1}{2}at^2}}$

$\large{\tt \leadsto x = \dfrac{1}{2} \times 9.8 \times (4)^2}$

$\large{\tt \leadsto x = \dfrac{1}{2} \times 9.8 \times 16}$

$\large{\tt \leadsto x = \dfrac{1}{\cancel{2}} \times 9.8 \times \cancel{16}}$

$\large{\tt \leadsto x = 1 \times 9.8 \times 8}$

$\large{\tt \leadsto x = 1 \times 78.4}$

$\huge{\boxed{\boxed{\tt x = 78.4 \: meters}}}$

So, they meet 78.4 meters from the top of the tower.

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