Answer:

We just need to find out the distance travelled by both the balls and then simply have to add them and problem will get solved :)

let’s see how to do it.



here I have represented the situation in a diagram.

height of the tower is AB = 200 m

Initial velocity of a ball thrown from the tower is = 0 m/s

and that of the ball thrown from the ground is = 100 m/s

‘P’ is the point at which both the ball crosses each other, let say time was ’t’ second.

Let distance covered by the ball thrown from tower in time t be S1S1 and that of the ball thrown from the ground be S2S2

Now, we will just find the distances S1S1 and S2S2 . sum of these distances will be equal to 200 m.

so, from second equation of motion, distance S1S1 travelled by the first ball will be

S1=12⋅g⋅t2S1=12⋅g⋅t2

and distance S2S2 travelled by ball from ground will be

S2=u⋅t−12⋅g⋅t2S2=u⋅t−12⋅g⋅t2

now, S1+S2=200mS1+S2=200m

substituting the values

12⋅g⋅t2+u⋅t−12⋅g⋅t2=200m12⋅g⋅t2+u⋅t−12⋅g⋅t2=200m

on simplifying we will get

u⋅t=200u⋅t=200 m

t=200100t=200100

t=

Answer:

Explanation:The simplest reference point for this kind of problem is the place where the fall begins, and to treat “down” as positive relative to it, since all motion will be downward.

The applicable SUVAT formula for uniformly-accelerated velocity versus time is

v = u +at

where v is the changing velocity, u is the initial velocity (0 for an object dropped from rest), a is the acceleration (g in this case), and t is time since acceleration began.