Vertical Velocity
A stone is thrown vertically upwards so as to reach 10 meters height in 10 seconds.
Now the question is not so simple, think! the velocity, with which the stone should be thrown to attain the same height in half the time will be more than the first velocity or less than the first velocity? What will be the required velocity?
Now the question is not so simple, think! the velocity, with which the stone should be thrown to attain the same height in half the time will be more than the first velocity or less than the first velocity? What will be the required velocity?
Labels: mathschallenge
posted by Rajesh Lal at
3:03 AM
11 Comments:
Double the power that was used 2 throw the stone up originally
I would just like to give a health and safety warning, please be careful not to stand directly below the stone after it has been throw up when trying to find the answe to this question.
Regards Windy (with a bump on his noggin)
2 meters per second
'cos the first one was 1mps
i wish i could have read the windymiller's warning before, almost broke my nose :)
Nice one...the answer is that if u want to throw the stone to exactly 10 meters in half the time, then its not possible. There is ONE and only ONE velocity at which the stone can reach a height of 10 meters in 10 seconds, if u increase the speed to a higher value then the stone will reach the height of 10 meters in half the time, but WILL NOT stop at that height and will continue climbing.
For every combination of time and speed, there is only one value of speed possible. If one were to change the value of time and speed, then the distance traveled cannot be the same and will also change.
Just one more clarification, the above answer will apply only in case of a body acted upon at all times by a uniform acceleration (in this problem, acceleration = gravity = 9.8 m/s/s)and NOT in the case of moving bodies with uniform speed.
If you throw something up in the air and after 10 seconds of being in the air it is 10 meters up, I would presume that this is because it is falling to the ground.
Therefore to attain the same height in half the time you need to throw the stone with less velocity.
Bigly
quick thought, if you were to throw the stone in such a way at a given higher velocity in which case it would reach a higher point but then fall and as it passes the 10m mark the time elapsed would be half the original time, or is this impossible due to frequency laws like a pendulum?
The motion is governed by simple kinematics, which is learned in a first year physics course. All the usual disclaimers apply - we assume no air resistance, uniform gravity, etc., which works surprisingly well at reasonably small heights and velocities for a dense object, like a rock.
For height h at time t, we have:
h(t) = v0*t - g*t^2/2
v0 is the initial velocity and g is the gravitational constant, 9.81 m/s^2 or so.
For the stated conditions, h(10)=10, substituting into the above and solving for v0, we get v0 = 50.05
If we do the same for h(5) = 10, we get v0 = 26.53
So, that's the answer, but what in the world does it mean? How can we get there in half the time by going at about half the speed? Seems ridiculous. So let's do a bit of exploring to see what this means.
From the statement of the problem the model is not entirely clear:
- is the stone at 10m as the highest point of it's flight
- otherwise, is it at 10m on the way up or down
Assume that 10m is the highest point in the flight. That means the velocity v at that point is zero.
But,
v(t) = v0 - g*t,
which you can remember from your physics, or get by differentiating h(t) once (or just think: for every second of flight the speed gets reduced by acceleration of gravity g, until it is reduced to zero, after which it increases in the negative direction, meaning the stone is going down).
When v(t) = 0, the above gives t = v0/g. If we put this back into the formula for h(t), we get:
h = v0^2/g - 1/2 * g * v0^2/g^2
Solving for v0, we get v0 = sqrt(2*g*h); that's the initial velocity required to reach height h at the top of the flight. For h = 10, this is about 14.1 and happens at t = 14.1/g, around 1.4 sec.
Conclusion: if the stone is thrown slower, than 14.1 m/s, it will not reach 10m, and otherwise, it will get there in 1.4 sec or faster on the way up. We now know that the problem describes a case where the stone is at 10m height in 10 sec on the way _down_, and so it makes a lot of sense that to get it there faster, it has to spend less time in the air, therefore we must throw it slower.
I do not recall seeing a problem before, where it took more time to understand the result, than to calculate it. This was a lot of fun.
Thanks, Rajesh - keep them coming!!!
P.S. For those curious about the disclaimers about various assumptions at the beginning, I recently analyzed the scenario for hurled pumpkins at high velocity (see the punkin chunking site for details) and wrote a simulator for that; air resistance definitely could not be ignored there.
Aha!
he question does not state that ten meters is the highest point for the faster throw.
;-)
There are a few options because of the vague question:
1) the stone is at the top of its throw at 10m, in which case it is launched at 14m/s.
2) if it's launched slower, it never reaches 10 m in height.
3. if faster, then it must be launched at 50 m/s. Gravity decelerates it at -9.8 m/s/s, of course. Changing the time from 10 to 5 seconds, you would think you need to throw the stone faster, but becuase it's not at 10 m on the way up, but rather the way down, you must actually throw it slower. Gravity doesn't change and still decelerates it at the same rate of -9.8 m/s/s, but for it to go up and back down in less time, it can't go as high, so it must be launched at a slower speed. The uniform acceleration formula in physics that is used is:
d = v1t + (1/2)at^2
Using this again, we find that the speed must be 26.5 m/s.
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