# Easy Time, Speed & Distance Solved QuestionAptitude 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.

## (8) Comment(s)

Jazzi
()

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

Vejayanantham TR
()

How come you take both takes same "t" ?

Rishi
()

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
()

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
()

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

Shahid
()

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
()

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

Krish
()

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