Aptitude Discussion

Q. |
A tower is 61.25 m high. A rigid body is dropped from its top and at the same instant another body is thrown up-wards from the bottom of the tower with such a velocity that they meet in the middle of the tower. The velocity of projection of the second body is: |

✔ A. |
24.5 m/s |

✖ B. |
20 m/s |

✖ C. |
25 m/s |

✖ D. |
22 m/s |

**Solution:**

Option(**A**) is correct

Let the body moving down-wards take '$t$' sec to reach half the height $\left(= \dfrac{61.25}{2} \right)$.

As we know,

$s=ut+\dfrac{1}{2}at^2$

Since, body is DROPPED from the top, initial velocity $u=0$. and $s=\dfrac{61.25}{2}$

\(\Rightarrow \dfrac{61.25}{2}=0+\dfrac{1}{2}\times 9.8\times t^2\)

\(\Rightarrow t=\dfrac{5}{2}\) sec

Again, assume that the second body is projected minimum velocity '$u$' up-wards, using the same formula,

$s=ut-\dfrac{1}{2}at^2$

\(\dfrac{61.25}{2}=u\times \dfrac{5}{2}- \left[\dfrac{1}{2} \times 9.8 \times \left( \dfrac{5}{2}\right)^2 \right]\)

\(\Rightarrow 61.25=u\times \dfrac{5}{2}- \left[\dfrac{245}{8} \right]\)

\(\Rightarrow 61.25=u\times \dfrac{5}{2}- \dfrac{61.25}{2}\)

\(\Rightarrow u=\dfrac{49}{2}\)

$= \textbf{24.5 m/sec}$

**Edit:** Taking note of the comments, solution has been updated.

**Jazzi**

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When the body falls from the top or moves towards the ground then gravitational force occurs on the body as an acceleration(9.8m/s/s).

When the body thrown upward from the ground then gravitational force occurs on the body as a deceleration.(- 9.8 m/s/s )

so, by default g=9.8 m/s/s

i.e the formula is changed over here for both the situation

**Vejayanantham TR**

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How come you take both takes same "t" ?

A rigid body is dropped from the tower's top and at the same instant another body is thrown upwards from the bottom of the tower with such a velocity that they meet in the middle of the tower.

So both the bodies spend the same time in the travel, making 't' same for both of the bodies.

**Suhail Ahmed**

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It is a Physics problem. The formula stated is derived of Netwon's Law of Gravity. And secondly it states free fall. I had forgotten the formula. Couldn't solve it.

**Fara**

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i also don't understand from where 245 came? n second thing, when a body is moving upward, 9.8 should b negative...

To your second query, Formula is used as $s=ut-\dfrac{1}{2}at^2$. So no need to take acceleration a negative value.

**Swathik**

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How come does this 245 took place.......?

**Krish**

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How is this calculation done.. can some one explain how 245 came to picture...

From where they got 9.8? i can't understand plz guide me