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There look to be a few problems. For the first questoin
In the vertical direction where gravity works, you have
* the initial velocity (resolved, which is positive)
* the acceleration (-g)
* the final distance (-5)
You should be able to use suvat to calculate the time, then plug that into the resolved horizontal velocity to calculate the horizontal distance.
Try that and post your reworking?
In the vertical direction where gravity works, you have
* the initial velocity (resolved, which is positive)
* the acceleration (-g)
* the final distance (-5)
You should be able to use suvat to calculate the time, then plug that into the resolved horizontal velocity to calculate the horizontal distance.
Try that and post your reworking?
Original post by ddsizebra
could someone assist me please?
I can't seem to make sense of this question and I'm not using the 5m as given in the question...
I can't seem to make sense of this question and I'm not using the 5m as given in the question...
Original post by mqb2766
There look to be a few problems. For the first questoin
In the vertical direction where gravity works, you have
* the initial velocity (resolved, which is positive)
* the acceleration (-g)
* the final distance (-5)
You should be able to use suvat to calculate the time, then plug that into the resolved horizontal velocity to calculate the horizontal distance.
Try that and post your reworking?
In the vertical direction where gravity works, you have
* the initial velocity (resolved, which is positive)
* the acceleration (-g)
* the final distance (-5)
You should be able to use suvat to calculate the time, then plug that into the resolved horizontal velocity to calculate the horizontal distance.
Try that and post your reworking?
do I also include that V for height is 0m/s?
No, that was one of the mistakes.
Vertically, v would only be zero at the top of the arc (maximum value of s). The vertical velocity is not zero when you enter the water.
You don't have to use this in the first part of the question.
Vertically, v would only be zero at the top of the arc (maximum value of s). The vertical velocity is not zero when you enter the water.
You don't have to use this in the first part of the question.
Original post by ddsizebra
do I also include that V for height is 0m/s?
(edited 4 years ago)
Original post by mqb2766
No, that was one of the mistakes.
Vertically, v would only be zero at the top of the arc (maximum value of s). The vertical velocity is not zero when you enter the water.
You don't have to use this in the first part of the question.
Vertically, v would only be zero at the top of the arc (maximum value of s). The vertical velocity is not zero when you enter the water.
You don't have to use this in the first part of the question.
Did you put 60 for the angle not 40?
Original post by old_teach
Did you put 60 for the angle not 40?
damn...sorry I'll correct it, I was doing another question and it got stuck in my head
Again a few mistakes. A key thing is to be clear about the horizontal and vertical motions. Gravity is vertical only.
The v^2 = u^2+2as gives the final velocity in the vertical direction. Did you combine with the horizontal to get the final speed?
You have to work out the time in the vertical direction. Why not use
s = ut+at^2/2
Then use the time to get the horizontal distance.
The v^2 = u^2+2as gives the final velocity in the vertical direction. Did you combine with the horizontal to get the final speed?
You have to work out the time in the vertical direction. Why not use
s = ut+at^2/2
Then use the time to get the horizontal distance.
Original post by ddsizebra
Original post by mqb2766
Again a few mistakes. A key thing is to be clear about the horizontal and vertical motions. Gravity is vertical only.
The v^2 = u^2+2as gives the final velocity in the vertical direction. Did you combine with the horizontal to get the final speed?
You have to work out the time in the vertical direction. Why not use
s = ut+at^2/2
Then use the time to get the horizontal distance.
The v^2 = u^2+2as gives the final velocity in the vertical direction. Did you combine with the horizontal to get the final speed?
You have to work out the time in the vertical direction. Why not use
s = ut+at^2/2
Then use the time to get the horizontal distance.
it did do all that but now I've corrected the angle but still wrong answer... in fact it's further away from the actual answer given
First calculation (you seem to have sq rooted a - number!!), s should have been -5 m
Original post by old_teach
First calculation (you seem to have sq rooted a - number!!), s should have been -5 m
it doesn't matter if a negative is square rooted as a number squared will always cause a positive value
Original post by ddsizebra
it doesn't matter if a negative is square rooted as a number squared will always cause a positive value
Not true! A negative number square rooted gives an imaginary number, which is fun.
If you're doing a calculation, and you get to try to sq root a negative, you know you've made a mistake, which is helpful.
Please try with s = -5!
Gravity does not operate horizontally.
And I gave you the formula to get time, pls use it.
Edit - read properly and you are using it in the vertical direction. The wording threw me.
And I gave you the formula to get time, pls use it.
Edit - read properly and you are using it in the vertical direction. The wording threw me.
Original post by ddsizebra
it doesn't matter if a negative is square rooted as a number squared will always cause a positive value
(edited 4 years ago)
Original post by old_teach
Not true! A negative number square rooted gives an imaginary number, which is fun.
If you're doing a calculation, and you get to try to sq root a negative, you know you've made a mistake, which is helpful.
Please try with s = -5!
If you're doing a calculation, and you get to try to sq root a negative, you know you've made a mistake, which is helpful.
Please try with s = -5!
My velocity is 10.1m/s now but S=1.912m
Final velocity in the vertical direction is negative.
Original post by ddsizebra
My velocity is 10.1m/s now but S=1.912m
OK so v = - 10.1 m/s is the correct value for vertical velocity, but you need to use pythag to add horiz vel to get actual velocity and work out angle (use tan?).
As mqb2766 says v is negative, which will make your answers work out
As mqb2766 says v is negative, which will make your answers work out
Original post by mqb2766
Final velocity in the vertical direction is negative.
I got it as a positive since acceeration and distance was minus, time 2 minuses makes a positive
Original post by ddsizebra
I got it as a positive since acceeration and distance was minus, time 2 minuses makes a positive
No, you were square rooting. There are two solutions, one with v going up (+) and the one you need, when v is going down (-).
Original post by ddsizebra
My velocity is 10.1m/s now but S=1.912m
Start again - DEEP BREATH
(a) use s = ut + 0.5at^2 to get t
then subst that t into s = ut for horizontal distance
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