Question

A stone falls from a height of 200m. Simultaneously another stone is thrown upwards with a velocity of 20m/s. Calculate when and where the 2 stones will meet?

Answer 1

Answer:

Explanation:

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We Height of = 200m

Initial velocity =50 m/s for stone b

Initial velocity for stone a = 0

Free fall from 200 m - stone a

Upward projection from group - stone b

Taking both cases simontanously.

Let after time ‘t’ both meets each other

Let distance travelled by - Stone a =x m

Then by ( stone b)= 200-x

Using the 2nd equation of motion for ( stone a)

S =ut+1/2at^2

X=0*t+1/2(10)t^2

Using 2nd equation of motion for ( stone b)

S = ut+1/2at^2

200-x = 50t+1/2(-10)t^2

We have 2 equations and 2 variables system

So by solving 1&2

1: x=5t^2

2: 200-x=50t-5t^2

We get

t =4s and x = 80 m

So they meet at 120 m from ground after a time of 4 sec.