Question

From top of a building of height 100 m an object is dropped , at a same time another object is thrown from the ground with speed 20 m/s . Find the position from the ground and time where these two objects meet each other

Answer 1

We will use the concept of relative velocity because the acceleration of both the objects are same and equal to "g"

Initial velocity of dropped object = 0 m/s

Initial velocity of thrown object = 20 m/s

Relative velocity of one object with rest to another = 20 m/s

Distance between the two objects = 100 m

Time taken = t

$t = \frac{Distance}{speed}$

$t = \frac{100}{20}$

t = 5 seconds

Hence, the time at which they will meet is 5 seconds

Now, let us find the distance covered by the dropped ball in 5 seconds

$s = ut + \frac{1}{2}\times a \times t^2$

$s = 0 + \frac{1}{2} \times 10 \times 5^2$

s = 125 m

Since, 125 m is greater than the height of the building. Hence, they will meet at the ground.

Hence, the position is at ground.

And the time at which they will meet will be 4 seconds