# Free falling object #2

an object, P is thrown vertically up and after 3s later another object ,Q is thrown up with the same initial speed of 20m/s and at the same position. find:
a) the time, answer: [T][/P]= 3.54s, [T][/Q]= 0.54s

2. i use this formula v=u + at
v=0 , u=20m/s a=9.8 ,t=?

anyone helps me~thank you

## The Attempt at a Solution

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Sorry but I'm having a hard time understanding the question. Could you write the question out in full? What time are you referring to?

find the time for object p and time for object Q

answer is time for object p is 3.54s， time for object Q is 0.54s

For what positions of P and Q, you want the time?
Like, do you want the time when the body reaches the ground or somewhere else?
Specify the positions and correct your question. I don't understand what you have written like this:-
[T][/P] and [T][/Q].

Object P is going to reach a maximum height and then fall back down to the ground again.

Object Q will be thrown 3 seconds later, at which point Object P will be returning to the ground.

I believe he wants to find the point in time at which Object P and Object Q are at the same height above the ground.

Object P is going to reach a maximum height and then fall back down to the ground again.

Object Q will be thrown 3 seconds later, at which point Object P will be returning to the ground.

I believe he wants to find the point in time at which Object P and Object Q are at the same height above the ground.

I think you are right because when i tried to solve it as you said, i got the answer which stupif has given in the main post.

It would go like this:-
(I am taking g=10m/s2 for easy calculation)

Since you want to find out the time when both the bodies are at same position, therefore there displacement is same. Let the displacement of Object P be s1 and Object Q be s2.

(initial velocity is denoted by u, here accceleration is negative since g is always downwards.)
Therefore,
s1 = s2
$\Rightarrow$ut-$\frac{1}{2}$at2=u(t-3)-$\frac{1}{2}$a(t-3)2

Plugin the values and solve, you will get t=3.5s. (Since i have taken g=10m/s2)
Solve for t-3 and you will get 0.5s.
Done..!!

sorry....that is the information i had....
and thank you, i got it.
but i wonder why my solution cant get the answer?
can someone explain to me why this formula cant use, v=u + at

sorry....that is the information i had....
and thank you, i got it.
but i wonder why my solution cant get the answer?
can someone explain to me why this formula cant use, v=u + at

We can't use the formula v=u+at becuase both the objects doesn't reach their max height in 2.04s. Object Q is thrown three seconds later. Let's visualise:-

Object P is thrown and attains the max. height in 2.04s. But still object Q is not thrown. Now the object P comes downwards and after 0.96s, Object Q is thrown upwards. Now object P is coming downwards and Object Q is going upwards. At one point they will meet each other and at that point their Displacement is going to be same.

now the second question is to find the velocity of P and Q. answer is velocity of P is -14.7m/s, velocity of Q is 14m/s

i use this formula v=u +at
but i cant get the answer.
v=?
u= 20m/s
a=9.8m/s2
t=3.54s

now the second question is to find the velocity of P and Q. answer is velocity of P is -14.7m/s, velocity of Q is 14m/s

i use this formula v=u +at
but i cant get the answer.
v=?
u= 20m/s
a=9.8m/s2
t=3.54s

Now we have got the time for both the objects, so you need to plugin the values of Object P and Object Q in the formula v=u+at.

Why you didn't get the answer? Did you take care of the signs?

ya. i got it. sorry for asking such a easy question.....and thank you very much

ya. i got it. sorry for asking such a easy question.....and thank you very much