Is it constant?
A stone is thrown vertically upward with an initial speed of 2 8 m s − 1 . Let it's velocity one second before it reaches the maximum height be V m s − 1 . Does the value of V change if the initial speed was increased?
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1 solution
We may think of it in another way. The upward journey is indical to the downward journey. The speed of the particle at the start of the last second of the upward journey is equal to the end of the first second of the downward journey.
So, if we start from the top and calculate the speed after one second it will be same for all the particles irrespective of their initial speed of throw and that will be
v = 0 + g ( 1 )
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Yeah, I tried to say that in my last sentence as well.
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Oh, yes. Pardon me, I skipped that line in my first reading.
Hey Steven @Steven Chase Your solution is good . However you really don't need calculus You can show that the value of V is always g i.e. 9.8
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Yeah, that's the same as my g Δ t , right? If you plug 1 in for Δ t . And Δ t is not meant to be an infinitesimal. It's just a generalization of your problem.
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Oh ya thanks
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@A Former Brilliant Member – Ah, but if we made it "an hour before it reaches the maximum height", then the answer changes!
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@Calvin Lin – Yes that's true
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@A Former Brilliant Member – It is mentioned in the question that the initial speed of the stone is 2 8 m s − 1 . Hence, one hour before reaching the maximum height, the stone would be in the hands of the thrower or lying somewhere on the ground. I don't think it is justified to ask this.
If you are suggesting that the particle is thrown at a great speed that it will take much more time to reach the maximum height then falling through that great altitude for a long time the stone will pass through a variable gravity. Therefore, the answer would be v = ∫ 0 Δ T g h d t instead of g surface Δ t .
If the time is increased to one hour then the initial speed will be so high that the particle will escape and never reach its maximum height.
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@Rohit Gupta – With the edit of "an hour before it reaches maximum height", the implicit assumption is that we choose a high enough initial velocity for the stone to travel for (say) 4 hours.
The issue that I'm raising here, is that g ≈ 9 . 8 is no longer accurate, and we thus have to start using calculus to figure out what the speed would be since acceleration is not constant.
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@Calvin Lin – I agree with you. It should be mentioned in the problem that the acceleration due to gravity remains constant.
@Calvin Lin – Yes, indeed. We must generalize within limits :)
I have clarified the meaning of Δ t in the original solution as well.
Call the initial speed v 0 . The time at which maximum height is reached is t f = g v 0 . Consider some intermediate time before that at which the speed is measured (in this problem, Δ t = 1 ) .
t i = t f − Δ t = g v 0 − Δ t
The speed at the intermediate time is:
v i = v 0 − g t i = v 0 − g ( g v 0 − Δ t ) = v 0 − v 0 + g Δ t = g Δ t
Thus, there is no dependence on the initial speed. Note that we could have also started at the top and then calculated the speed after a duration of Δ t , which would have also given us a result of g Δ t . In this problem, g Δ t = g since Δ t = 1 .