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# Can anyone help out with this mechanics-type question? watch

1. I'm working through a revision pack, and came across a question which I'm struggling to get right. I know the theory and the steps behind answering it, but am unsure of why I keep getting the wrong answer. Can anyone help?

The question:

A bullet is fired vertically upwards at a speed of 147m/s. Find the time for which the bullet is at least 980m above the level of projection.

The acceleration is always going to be 9.81 (or -9.81) as the question is in the context of free fall under gravity.

Thanks a bunch guys!
2. Is the answer t = 13.345... seconds ?

Let me know if its correct. I'll guide you through.
3. (Original post by TurgidMeniscus)
I'm working through a revision pack, and came across a question which I'm struggling to get right. I know the theory and the steps behind answering it, but am unsure of why I keep getting the wrong answer. Can anyone help?

The question:

A bullet is fired vertically upwards at a speed of 147m/s. Find the time for which the bullet is at least 980m above the level of projection.

The acceleration is always going to be 9.81 (or -9.81) as the question is in the context of free fall under gravity.

Thanks a bunch guys!
Use one of the SUVAT equations relating time, initial velocity, acceleration and displacement. Then rearrange for t to find the time. Note that the acceleration should be negative as the ball will decelerate as it will go upwards.
4. (Original post by Mehrdad jafari)
Use one of the SUVAT equations relating time, initial velocity, acceleration and displacement. Then rearrange for t to find the time. Note that the acceleration should be negative as the ball will decelerate as it will go upwards.
Thanks! I was wondering, given that velocity is a vector quantity, would you need to use a negative v value? No forces are at work, so how would you even determine the positive/negative direction?
5. (Original post by TurgidMeniscus)
Thanks! I was wondering, given that velocity is a vector quantity, would you need to use a negative v value? No forces are at work, so how would you even determine the positive/negative direction?
That's a good point! The Mathematical sign is to indicate the direction with respect to the other quantity being affected by it, velocity in this case, but because the velocity has already been stated as positive then the acceleration will be negative as it's in the opposite direction to initial velocity.

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6. Listing the components of suvat that we know, we can use the eqn s=ut+0.5t^2

Substitute in the values a = -9.81, u = 147, and s = 980 (since s is AT LEAST above 980, we can use 980m as the lowest boundary).

You should get a quadratic which is 0 = -4.905t^2 + 147t - 980
This is solvable using plysmlt2 on a GDC - you should get t = 10.0102... and t = 19.95918...
I'm guessing that the answer is 10.0s as it's the smallest positive value that t can be.

Hope this helped Tell me if it's the right answer!
7. (Original post by anamariamikhail)
Listing the components of suvat that we know, we can use the eqn s=ut+0.5t^2

Substitute in the values a = -9.81, u = 147, and s = 980 (since s is AT LEAST above 980, we can use 980m as the lowest boundary).

You should get a quadratic which is 0 = -4.905t^2 + 147t - 980
This is solvable using plysmlt2 on a GDC - you should get t = 10.0102... and t = 19.95918...
I'm guessing that the answer is 10.0s as it's the smallest positive value that t can be.

Hope this helped Tell me if it's the right answer!
Why have you kept the negative -4.905^2? You could just add it to the other side, to get 4.905t^2 - 147t + 980. Then your 2 values for t are the 2 times when the bullet passes 980m. You take the times away from each other to find the time that it is above those 2 points.
8. (Original post by richpanda)
Why have you kept the negative -4.905^2? You could just add it to the other side, to get 4.905t^2 - 147t + 980. Then your 2 values for t are the 2 times when the bullet passes 980m. You take the times away from each other to find the time that it is above those 2 points.
It's a personal thing... It gives the same answers though. The last part makes a lot of sense, was a little confused about choosing between the numbers myself. Thanks
9. (Original post by anamariamikhail)
Listing the components of suvat that we know, we can use the eqn s=ut+0.5t^2

Substitute in the values a = -9.81, u = 147, and s = 980 (since s is AT LEAST above 980, we can use 980m as the lowest boundary).

You should get a quadratic which is 0 = -4.905t^2 + 147t - 980
This is solvable using plysmlt2 on a GDC - you should get t = 10.0102... and t = 19.95918...
I'm guessing that the answer is 10.0s as it's the smallest positive value that t can be.

Hope this helped Tell me if it's the right answer!
I wasn't actually thinking about the quadratic form of the equation. I guess it would be better to find the final velocity and then use the equation v=u+at to find the time
10. Is my method wrong?

Well as, we know final velocity is zero at the top of the bounce, so we can find the time taken for the ball to reach the top of the bounce, with the equation v=u+at, 0=147-9.81t, which would give 14.98 s

Now we put the 14.98s along with other values into the equation s = ut+1/2at^2 to find the distance covered till top of the bounce

which is s = 147*14.98 + 1/2 * -9.81 * (14.98)^2

s = 1101m~

So as it takes, 14.98 secs to cover 1101m vertically, we can also take ratios and say, it takes 14.98/1101 = 0.0136... seconds to cover 1m, and hence multiply by 980, which would give 13.32 s approx~

Please let me know what's wrong in this, I also actually got the quadratic equation before, but figured it's probably out of the scope of High school projectiles or so lol.
Is my method wrong?

Well as, we know final velocity is zero at the top of the bounce, so we can find the time taken for the ball to reach the top of the bounce, with the equation v=u+at, 0=147-9.81t, which would give 14.98 s

Now we put the 14.98s along with other values into the equation s = ut+1/2at^2 to find the distance covered till top of the bounce

which is s = 147*14.98 + 1/2 * -9.81 * (14.98)^2

s = 1101m~

So as it takes, 14.98 secs to cover 1101m vertically, we can also take ratios and say, it takes 14.98/1101 = 0.0136... seconds to cover 1m, and hence multiply by 980, which would give 13.32 s approx~

Please let me know what's wrong in this, I also actually got the quadratic equation before, but figured it's probably out of the scope of High school projectiles or so lol.
It is not mentioned that the maximum height is 980, which in fact is 1101m as you worked it out using the other equation.

You could use v2=u2+2as to find the final velocity and then use v=u+at to find the time to reach the specified height
12. (Original post by Mehrdad jafari)
It is not mentioned that the maximum height is 980, which in fact is 1101m as you worked it out using the other equation.

You could use v2=u2+2as to find the final velocity and then use v=u+at to find the time to reach the specified height
So i guess taking ratios is just wrong in projectiles?
So i guess taking ratios is just wrong in projectiles?
That could possibly be a good method if there was no change in velocity but because the velocity is constantly changing you cannot use ratios. For example, if there was no gravity to decelerate the ball, then you would intuitively know that if, for example, it took a ball 10s to travel 50 meters, it would definitely take 20s for it to travel 100 meters.
14. (Original post by Mehrdad jafari)
That could possibly be a good method if there was no change in velocity but because the velocity is constantly changing you cannot use ratios. For example, if there was no gravity to decelerate the ball, then you would intuitively know that if, for example, it took a ball 10s to travel 50 meters, it would definitely take 20s for it to travel 100 meters.
Right, Thank you.

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