Why Do Objects P and Q Have Different Times to Reach Their Peak?

[stupif]

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=?
but my answer is 2.04s

anyone helps me~thank you

The Attempt at a Solution

 

[nayfie]

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?

 

[stupif]

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

 

[Saitama]

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].

 

[nayfie]

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.

 

[Saitama]

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..!

 

[stupif]

sorry...that is the information i had...
and thank you, i got it.
but i wonder why my solution can't get the answer?
can someone explain to me why this formula can't use, v=u + at

 

[Saitama]

sorry...that is the information i had...
and thank you, i got it.
but i wonder why my solution can't get the answer?
can someone explain to me why this formula can't 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.

 

[stupif]

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 can't get the answer.
v=?
u= 20m/s
a=9.8m/s2
t=3.54s

 

[Saitama]

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 can't 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?

 

[stupif]

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

 

[Saitama]

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

Glad you got it! And your welcome!