a stone is thrown vertically upward with velocity 19.6 m/s . after 2second another stone is thrown upward with velocity 9.8m/s. when where will these stone collide?(g=9.8m/s2)
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Well, there is an easy way to solve any such question. Just follow the steps given!
So here we go!
Now, to make sure that we can use the formulae in a much easier fashion, and also to avoid confusing ourselves, let's define a coordinate system. My choice is the point from where you throw the balls as the origin and the upwards direction as the positive direction.
Clearly from the above diagram, the acceleration due to gravity ( coincidently the only acceleration the body has) is negative, while the initial velocity and position are positive. You will see why I have put position in bold in a while.
Now we know the equations of motion( which can be used as the acceleration is constant). Let us use them.
We know that the positions of the object are the same when they meet. So we can equate the positions. This is why I had put position in bold : because the h term is commonly mistaken to mean displacement.
We can see that the time of ascent is 2 s. So when they meet, they meet on the downward journey of the first body. So the sum of the displacements should be the maximum height reached by the first ball.
Now all you have to do is substitute the values in the equation and equate the positions to get the time taken by the first body. I'm leaving this part for you to do. Good luck.
Hope this helped.
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Muthusamy P, M.Sc. Retired as Head of Department of Physics. Government of Tamilnadu, lndia.
Answered Sun · Author has 1.4k answers and289.2k answer views
Acceleration due to gravity is g = 9.8 m/ s^2.
Therefor the velocity of the first thrown stone will reduce by 2*9.8 = 19.6 m/s in 2 seconds.
That is it's velocity is zero after 2 second.
The height to which it has gone up is average velocity * time = 9.8 * 2 = 19.6 m
Now the second stone is thrown up.
Since both are subjected to the same g, we can neglect it and consider the motion as if there is no gravity.
The first stone is at rest at a height of 19. 6 m from the ground and the second one moves up with a velocity of 9.8 m/s .
Time to reach the first ball is distance / velocity = 19.6 / 9.8 = 2 second.
In 2 second , the second one meets the first one from the time it was thrown up.
Totally in 4 second from the start of first stone.
To find where do the meet,
The first ball reaches the ground in 2 second from the height of 19.6 m , since it has taken 2 s to move up.
Therefore both meet at the starting point.
We can use the second one too to find the meeting place.
In 1 s it's velocity reduces to zero and in another 1 s it reaches the ground.
Therefore both meet at the starting point.
So here we go!
Now, to make sure that we can use the formulae in a much easier fashion, and also to avoid confusing ourselves, let's define a coordinate system. My choice is the point from where you throw the balls as the origin and the upwards direction as the positive direction.
Clearly from the above diagram, the acceleration due to gravity ( coincidently the only acceleration the body has) is negative, while the initial velocity and position are positive. You will see why I have put position in bold in a while.
Now we know the equations of motion( which can be used as the acceleration is constant). Let us use them.
We know that the positions of the object are the same when they meet. So we can equate the positions. This is why I had put position in bold : because the h term is commonly mistaken to mean displacement.
We can see that the time of ascent is 2 s. So when they meet, they meet on the downward journey of the first body. So the sum of the displacements should be the maximum height reached by the first ball.
Now all you have to do is substitute the values in the equation and equate the positions to get the time taken by the first body. I'm leaving this part for you to do. Good luck.
Hope this helped.
7.8k Views · View Upvoters
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Muthusamy P, M.Sc. Retired as Head of Department of Physics. Government of Tamilnadu, lndia.
Answered Sun · Author has 1.4k answers and289.2k answer views
Acceleration due to gravity is g = 9.8 m/ s^2.
Therefor the velocity of the first thrown stone will reduce by 2*9.8 = 19.6 m/s in 2 seconds.
That is it's velocity is zero after 2 second.
The height to which it has gone up is average velocity * time = 9.8 * 2 = 19.6 m
Now the second stone is thrown up.
Since both are subjected to the same g, we can neglect it and consider the motion as if there is no gravity.
The first stone is at rest at a height of 19. 6 m from the ground and the second one moves up with a velocity of 9.8 m/s .
Time to reach the first ball is distance / velocity = 19.6 / 9.8 = 2 second.
In 2 second , the second one meets the first one from the time it was thrown up.
Totally in 4 second from the start of first stone.
To find where do the meet,
The first ball reaches the ground in 2 second from the height of 19.6 m , since it has taken 2 s to move up.
Therefore both meet at the starting point.
We can use the second one too to find the meeting place.
In 1 s it's velocity reduces to zero and in another 1 s it reaches the ground.
Therefore both meet at the starting point.
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