Question

A ball is dropped from the top of a building of
height 80 m. At same instant another ball is thrown
upwards with speed 50 m/s from the bottom of the
building. The time at which balls will meet is
(1) 1.6 s
(2) 5s
(3) 8s
(4) 10 s
The diagram is attached
plz answer​

Answer 1

Explanation:

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case-1. ball dropped

use second eqn of motion:

s=ut+1/2at²

let the height be h metres

u is 0

let t be the time when both objects meet

acc is 10

=> h=0+1/2(10)t²

=> h=5t²

case - 2 ball thrown

let the height be (h-80) metres

u = 50

acc = 10

(h-80)=(50)t+1/2at²

h-80=50t+5t²

replacing value of h

you will get the value of t as 1.6 seconds...

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Answer 2

I don’t now the answer