Question

11. A ball is dropped from a high rise platform at
t=0 starting from rest. After 6 seconds another
ball is thrown downwards from the same
platform with a speed y. The two balls meet at
t= 18 s. What is the value of v?
(Take g=10 m/s2)
(a) 75 m/s
(b) 55 m/s
(c) 40 m/s
(d) 60 m/s follow me ♥♥inbox me ♥​

Answer 1

Neither I'm going to follow you or Inbox you. This is not fbook.

Answer:

a) $v =75ms^{-1}$

Explanation:

pick a ball and drop it after 17 seconds it will cover distance

$s = ut + \frac{1}{2}at^{2}$

u = 0

a = 10 ms-2

t = 18

so put values and we have

$s = (0*18) + (\frac{1}{2}*10*18^{2})$

$s =1620m$

pick the other ball and through it downward with velocity v after 6 seconds

after (18-6) second it reaches 1st ball means both ball covered same distance in time t = 12

use same formula this time u = v

$s = ut + \frac{1}{2}at^{2}$

u = v

a = 10 ms-2

t = 12

so put values and we have

$1620 = (v*12) + (\frac{1}{2}*10*12^{2})$

$v =75ms^{-1}$

Answer 2

sorry mate

i dont know the answer of itt