You are Here: Home >< Physics

# As physics mechanics question (urgent) watch

1. from the new textbook pg 35
a net ball is thrown with a velocity of 6 m/s at am angle of 40 from the vertical, towards a netball hoop.
a) if the hoop is 0.9 m above the point of release, will it rise high enough to go in the hoop?
b) if the centre of the hoop is 3 m away horizontally from the point of release explain whether or not you believe this throw will score in the hoop. explain with calculations

from the new textbook pg 35
a net ball is thrown with a velocity of 6 m/s at am angle of 40 from the vertical, towards a netball hoop.
a) if the hoop is 0.9 m above the point of release, will it rise high enough to go in the hoop?
b) if the centre of the hoop is 3 m away horizontally from the point of release explain whether or not you believe this throw will score in the hoop. explain with calculations

Have you drawn a diagram for yourself?

Think, suvat:

For part a, you need to see whether the highest point the ball reaches is equal to, or greater than 0.9 m or not. Therefore you need to consider the ball's vertical motion, ie the vertical component of its velocity in your calculations.
You're looking for S, you can find out what value you need to use for U, the acceleration of the netball is -9.81 ms^-2, and what is the netball's velocity at its highest point? (When it has travelled S metres?)
Can you clarify what kind of help you need for part b?
from the new textbook pg 35
a net ball is thrown with a velocity of 6 m/s at am angle of 40 from the vertical, towards a netball hoop.
a) if the hoop is 0.9 m above the point of release, will it rise high enough to go in the hoop?
b) if the centre of the hoop is 3 m away horizontally from the point of release explain whether or not you believe this throw will score in the hoop. explain with calculations

For b) the vertical component will take time t1 to reach the hoop (coming down, after going up), and the horizontal component will take time t2 to travel the 3 metres horizontally. If it is to go in the hoop, t1 = t2.

To find t1, use s = ut + (at^2)/2, where s = 0.9, u = 6cos(40), and a = -g.

To find t2, a = 0 so you can use t = d/v, where d = 3 and v = 6sin(40).

I found the two times to be different, so it won't go in the hoop. (Unless it's a massive hoop )
4. (Original post by ombtom)
For b) the vertical component will take time t1 to reach the hoop (coming down, after going up), and the horizontal component will take time t2 to travel the 3 metres horizontally. If it is to go in the hoop, t1 = t2.

To find t1, use s = ut + (at^2)/2, where s = 0.9, u = 6cos(40), and a = -g.

To find t2, a = 0 so you can use t = d/v, where d = 3 and v = 6sin(40).

I found the two times to be different, so it won't go in the hoop. (Unless it's a massive hoop )
my teacher was saying this type of questin wont be asked unless it goes in. so shes saying it should go in but shell do the calculations and let us know again
5. (Original post by ombtom)
For b) the vertical component will take time t1 to reach the hoop (coming down, after going up), and the horizontal component will take time t2 to travel the 3 metres horizontally. If it is to go in the hoop, t1 = t2.

To find t1, use s = ut + (at^2)/2, where s = 0.9, u = 6cos(40), and a = -g.

To find t2, a = 0 so you can use t = d/v, where d = 3 and v = 6sin(40).

I found the two times to be different, so it won't go in the hoop. (Unless it's a massive hoop )
i see you alot on physics and maths im really having a hard time in alvels i wrote about it here http://www.thestudentroom.co.uk/show....php?t=3671857
any tips?
my teacher was saying this type of questin wont be asked unless it goes in. so shes saying it should go in but shell do the calculations and let us know again
I was just thinking that – maybe they want you to think that? I'm quite tired (really shouldn't be in the Physics forums); I might've made a mistake.

i see you alot on physics and maths im really having a hard time in alvels i wrote about it here http://www.thestudentroom.co.uk/show....php?t=3671857
any tips?
It's nice to be noticed. I'll have a look!
7. (Original post by Mehrdad jafari)
I believe you asked a similar question a few days ago so I assume I wasn't being helpful

For part a you can easily work out the maximum vertical displacement as you know that the velocity at which the netball has reached maximum displacement will be 0. Therefore, using v^2=u^2+2as.

No matter how far the netball will travel horizontally, if the maximum vertical displacement is not equal or greater than 0.9m then the netball will not be thrown into the hoop. See the attached file.
Attachment 470609

Posted from TSR Mobile
thank you very much!
thank you very much!
No worries! I think other posters have said the same thing, I just wrote down the solution.

Posted from TSR Mobile
9. (Original post by Mehrdad jafari)
No worries! I think other posters have said the same thing, I just wrote down the solution.

Posted from TSR Mobile
do you do chemistry as well :P AS level?
do you do chemistry as well :P AS level?
I did AS last year.

Posted from TSR Mobile
11. (Original post by Mehrdad jafari)
I did AS last year.

Posted from TSR Mobile
https://www.pearsonschoolsandfecolle...-1-SAMPLES.pdf

page 3
answers to exam style questions part
first question; why is the least count multiplied by 2 (0.005 x2) ?
isnt it just 0.005 /0.03 x 100
https://www.pearsonschoolsandfecolle...-1-SAMPLES.pdf

page 3
answers to exam style questions part
first question; why is the least count multiplied by 2 (0.005 x2) ?
isnt it just 0.005 /0.03 x 100
I'm not really sure to be honest, considering that I wasn't good at practicals also. I cannot see where the measurement would be taken twice for the percentage error to be multiplied by two.
13. (Original post by Mehrdad jafari)
I'm not really sure to be honest, considering that I wasn't good at practicals also. I cannot see where the measurement would be taken twice for the percentage error to be multiplied by two.
same here /:
14. (Original post by Mehrdad jafari)
I believe you asked a similar question a few days ago so I assume I wasn't being helpful

For part a you can easily work out the maximum vertical displacement as you know that the velocity at which the netball has reached maximum displacement will be 0. Therefore, using v^2=u^2+2as.

No matter how far the netball will travel horizontally, if the maximum vertical displacement is not equal or greater than 0.9m then the netball will not be thrown into the hoop. See the attached file.
Attachment 470609

Posted from TSR Mobile
just looked at this and i think you did this wrong. it's 40 degree from the vertical and the diagram my teacher drew was different will let you know tomorrow anyway..............
same here /:
just looked at this and i think you did this wrong. it's 40 degree from the vertical and the diagram my teacher drew was different will let you know tomorrow anyway..............
You are right, sorry, I just remembered from last time too! . I have been a bit inattentive these days.

In that case, the scenario gets a bit complicated as there will be some assumptions to be made. The netball does seem to exceed the required height of the hoop but assuming that the ball will have to be thrown to the center of the hoop (as this seems to be the case since the question specifies the distance from the point of release to the centre of the hoop), then you can work out the time taken for the netball to travel the maximum vertical distance (which would be above the hoop), and the time taken for the netball to travel the minimum horizontal distance, 3m. Comparing the time taken for both components, it should turn out that for the netball to be thrown into the hoop the vertical and horizontal time components would have to be almost equal. This is because if the horizontal time component is equal to the vertical time component, it means at that moment the ball is just above the centre of the hoop. We can make this assumption (I think) as the netball doesn't exceed the vertical height by a great distance.

Oh god! that was just horrible
17. (Original post by Mehrdad jafari)
You are right, sorry, I just remembered from last time too! . I have been a bit inattentive these days.

In that case, the scenario gets a bit complicated as there will be some assumptions to be made. The netball does seem to exceed the required height of the hoop but assuming that the ball will have to be thrown to the center of the hoop (as this seems to be the case since the question specifies the distance from the point of release to the centre of the hoop), then you can work out the time taken for the netball to travel the maximum vertical distance (which would be above the hoop), and the time taken for the netball to travel the minimum horizontal distance, 3m. Comparing the time taken for both components, it should turn out that for the netball to be thrown into the hoop the vertical and horizontal time components would have to be almost equal. This is because if the horizontal time component is equal to the vertical time component, it means at that moment the ball is just above the centre of the hoop. We can make this assumption (I think) as the netball doesn't exceed the vertical height by a great distance.

Oh god! that was just horrible
I did the first question and it definitely reaches sufficient height.

For the second question, I did what you said, and I think my answers were about a second apart. Might've made a mistake in this one though.
18. (Original post by ombtom)
I did the first question and it definitely reaches sufficient height.

For the second question, I did what you said, and I think my answers were about a second apart. Might've made a mistake in this one though.
Yeah, your method was correct! Though, doing the calculations now, my vertical and horizontal time components are 0.66s and 0.78s respectively ( I chose the greater time when solving the quadratic as the smaller value would give the time for the ball to travel 0.9 vertically without going any higher and coming back). The difference in time doesn't seem to be significant but I guess that would make a difference.
19. (Original post by Mehrdad jafari)
Yeah, your method was correct! Though, doing the calculations now, my vertical and horizontal time components are 0.66s and 0.78s respectively ( I chose the greater time when solving the quadratic as the smaller value would give the time for the ball to travel 0.9 vertically without going any higher and coming back). The difference in time doesn't seem to be significant but I guess that would make a difference.
I recognise those numbers.
20. (Original post by ombtom)
I recognise those numbers.
All good

Posted from TSR Mobile

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 21, 2015
Today on TSR

### Edexcel GCSE Maths Unofficial Markscheme

Find out how you've done here

### OCR AS Chemistry Exam Discussion

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE