ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: sin0001 on March 09, 2013, 08:40:06 pm
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There was this question I was trying to solve, from Checkpoints, went something like this:
Ball released at speed of 6.5 ms-1, at an angle of 30 degrees to the horizontal, from a height of 1.5 m. It's allowed to land on the ground, and the question asked for the time the ball spent in its flight. (Answer is approx. 0.98 sec)
First I found the time taken to reach the max. height, and doubled it, but this was incorrect. (Dunno why it doesn't work..)
So I found the time taken to reach max. height and added the time taken from max. height to landing on ground (had to use equations of motion twice)
I was wondering if there's a faster approach to this question, in which you don't need to use equation of motion twice.
Thanks in advance
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It doesn't work because by doubling the time you are finding the time that it takes for it to come back down to a height of 1.5 m. It still needs to fall that extra 1.5 m to hit the ground.
So the best way of doing it is probably, as you have done, split it into the two equations of motion and then add the two times you find.
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It doesn't work because by doubling the time you are finding the time that it takes for it to come back down to a height of 1.5 m. It still needs to fall that extra 1.5 m to hit the ground. So the best way of doing it is probably, as you have done, split it into the two equations of motion and then add the two times you find.
A crude diagram if you want to visualise it, i've taken the starting point to be zero, the max height reached to be h:
(http://i.imgur.com/xHMcUm6.png)
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But the landing speed will be same as the speed of release (6.5) right?
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As the object moves through that extra 1.5 m, it is still accelerating downwards due to gravity, so it will still be 'speeding up'. I.e. The landing speed will be greater than the speed of release. When the object gets to the stage that it is 1.5 m above the ground and moving towards the ground it will be travelling at the speed that it was released. (note this is assuming no air resistance e.t.c)
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ohh okay that makes sense.
I thought there would be a faster, like a one-step formula to find time of flight, because there were only a few marks available for this.
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ohh okay that makes sense.
I thought there would be a faster, like a one-step formula to find time of flight, because there were only a few marks available for this.
Actually wtf am I thinking. You can just use the equation of motion once, and it should work, setting s=-1.5 below the starting point. You'd get a quadratic and have to solve it, picking the right solution. Sorry... don't know wtf I was thinking.
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Ohhhh there you go, that makes more sense lol
Thanks :)
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Coming back to the same question:
When I use thr equations of motion to solve for time of flight, is 'a' supposed to be -10 or 10? (x is -1.5 m)
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Coming back to the same question:
When I use thr equations of motion to solve for time of flight, is 'a' supposed to be -10 or 10? (x is -1.5 m)
It should be -10ms-2, because if displacement downwards is negative, then velocity and acceleration downwards should always be downwards.
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Coming back to the same question:
When I use thr equations of motion to solve for time of flight, is 'a' supposed to be -10 or 10? (x is -1.5 m)
If you define up as positive then g (a) will be negative, if you define down as positive then g (a) will be positive. In this case since the distance is below the starting point and is negative, we've taken up as positive and so a would be negative. If you defined down as positive then you would have s=1.5 m and a=+9.8 m/s^2 (you'd also have to fix up the direction for the initial velocity).
Really its best to just stick to using one way that you're used to, and work from there, it just depends on how you define things.