Throwing A Stone (1 Viewer)

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Hello all, I'm a 15 year old student in year 11 in Tasmania, Australia. This was a question from my Physical Sciences class that I was a bit confused about. Any help would be greatly appreciated.

1. The problem statement, all variables and given/known data
A stone is thrown vertically into the air and caught at the same height. It leaves the thrower's hand at 29.4ms^-1.

1. Find the time taken for the stone to reach its maximum height.
2. The maximum height reached.
3. The time taken for the stone to return to the height from which it was thrown.
4. The stone's velocity as it hits the thrower's hand on its return.


2. Relevant equations
I have figured out so far that:
S=0
u=29.4ms^-1
a=-9.81ms^-1


3. The attempt at a solution
1. Is this just 29.4/9.81 ?
2. The above answer x 9.81?
3. Answer number one x2?
4. Would you use v^2=u^2 + 2as ?

Thanks in advance.
 
what are the equations of motions???

can u write those down? u already have three of the vars...just use those equations to calulate the rest of the vars.
 

Mentallic

Homework Helper
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2. Relevant equations
I have figured out so far that:
S=0
u=29.4ms^-1
a=-9.81ms^-1
s=displacement (of the stone). This is not zero. The displacement is how high into the air it goes before stopping and turning back down to earth.
So we don't know s, but what we do know is that when the stone reaches the top of its flight, its final velocity (v) is zero.

So now use an equation that relates u, v, a, s to solve for s.
The equations is: [tex]v^2=u^2+2as[/tex]

To find the time taken for the stone to reach max height, you can use (once you find s) [tex]s=ut+\frac{1}{2}at^2[/tex]
but more easily, you can use [tex]v=u+at[/tex]

I can tell you now that since these questions are assuming no forces except for gravity are acting on the stone, the time for the stone to reach the top of its flight and then to return back into his hand is the same, and the velocity at which it returns to his hand is the same as the velocity at which it was released. But you should show this with the equations.
 
the velocity at which it returns to his hand is the same as the velocity at which it was released. But you should show this with the equations.
Thanks for replying. Does this mean that final velocity(v)=initial velocity(u) which is 29.4ms^-1?
 

Mentallic

Homework Helper
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94
Yes. But use the equations you know to find the final velocity.
u=0 (it's stopped at the top of its flight)
s= (height above point of release)
a= (gravity)

Now use [tex]v^2=u^2+2as[/tex] and you should notice that the final velocity, v, is the negative of the intial velocity of the throw (since the rock is going back the other way).
 

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