Help with Mechanics Questions

Hey

I could do with some help with the following mechanics (Kinematics) questions, I'm a little confused on which formula to use with each:

1. A tennis ball is thrown vertically upwards and reaches a maximum height of 4.1m. Find the initial speed of the ball.

2. A stone is fired vertically upwards from a catapult at 12.1ms^-1.
The catapult is 2.3m above the ground.
Calculate the maximum height reached by the stone.

3a. A slate falls from the roof of a house. It passes the top of a window after 0.8s. It passes the bottom of the window after a further 0.4s.
Find the speed of the slate at the top of the window.
3b. How tall is the window?

4a. A coin is thrown vertically upwards with speed 8ms^-1 from a height of 1.5m above the ground.
Find the greatest height reached by the coin
4b. Calculate the time taken for the coin to reach the ground

These are the 5 formulas:

s = ut + 1/2 x at^2
s = vt - 1/2 x at^2
v = u + at
s = 1/2(u + v)t
v^2 = u^2 + 2as

Any efficient help will be greatly appreciated, as I need to understand this.

Thank-you. :smile:

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Reply 1

TenOfThem

Original post by FranticR

1. A tennis ball is thrown vertically upwards and reaches a maximum height of 4.1m. Find the initial speed of the ball.

s = ut + 1/2 x at^2
s = vt - 1/2 x at^2
v = u + at
s = 1/2(u + v)t
v^2 = u^2 + 2as

You know
s
v
a
and you want
u

There is only one you can use

Reply 2

TenOfThem

Original post by FranticR

2. A stone is fired vertically upwards from a catapult at 12.1ms^-1.
The catapult is 2.3m above the ground.
Calculate the maximum height reached by the stone.

s = ut + 1/2 x at^2
s = vt - 1/2 x at^2
v = u + at
s = 1/2(u + v)t
v^2 = u^2 + 2as

You know
u
a
v

and you need to find
s

There is only one you can use

Reply 3

TenOfThem

Original post by FranticR

3a. A slate falls from the roof of a house. It passes the top of a window after 0.8s. It passes the bottom of the window after a further 0.4s.
Find the speed of the slate at the top of the window.
3b. How tall is the window?

s = ut + 1/2 x at^2
s = vt - 1/2 x at^2
v = u + at
s = 1/2(u + v)t
v^2 = u^2 + 2as

(a)
You know
u
a
t

and need v

only one you can use

(b)
You know
u
a
t (in 2 cases)

and you want
s

only one you can use

Reply 4

TenOfThem

Original post by FranticR

4a. A coin is thrown vertically upwards with speed 8ms^-1 from a height of 1.5m above the ground.
Find the greatest height reached by the coin
4b. Calculate the time taken for the coin to reach the ground

s = ut + 1/2 x at^2
s = vt - 1/2 x at^2
v = u + at
s = 1/2(u + v)t
v^2 = u^2 + 2as

(a)
you know
u
a
v

and you want
s

(b)
you know
u
a
s

and you need
t

Reply 5

FranticR

Original post by mackemforever

Seriously?

How ****ing lazy are you?

You have given us the questions and literally written out all of the formulas that you need to solve them and you expect other people to put 2 and 2 together for you?

How about you get off your lazy ****ing ass and show you've put some effort in. Show what working you have tried, what exactly you don't understand, where in each question you're going wrong instead of being a lazy ****ing **** and expecting other people to do all of your work for you.

I've not come here to be lazy, I have come here for some help with the said questions, I did not expect anyone to write the answers out for me.

There is also no need to insult me like the way you just did. Its a maths forum for crying out loud...

Reply 6

ilovechemistry2

Hi guys, was just wondering if anyone could help me on this M1 question from the edexcel textbook:

Page 30 Exercise 2.4

Theresa is handing the baton onto Magda in a relay race. Theresa is running at a constant speed when she is 4.5m away, Magda starts running with acceleration 1ms^-2. Theresa continues at a constant speed and just manages to catch up and hand over the baton. How fast was Theresa running?

the answer in the back of the book is 3ms^-1 but i have no idea how you reach that answer - if anyone could show me a step by step method that would be very helpful

thanks in advance

Reply 7

mackemforever

Original post by FranticR

Asking for help is fine.

Asking for help when you have done absolutely nothing to show that you have even attempted the question is not.

Reply 8

davros

Original post by ilovechemistry2

Not wishing to be rude, but it's considered polite to start your own thread in which to ask your questions, rather than hijacking a thread started by someone else for their questions :smile:

Reply 9

davros

Original post by FranticR

Start with the first question, show us the working you have done so far and where you are stuck. Then someone can try to help you out :smile:

Reply 10

FranticR

Original post by mackemforever

Asking for help is fine.

Asking for help when you have done absolutely nothing to show that you have even attempted the question is not.

Still there was no need for your over-reaction. I'm new, I'm doing my best, you've not given me a very good welcome.

Reply 11

ilovechemistry2

Original post by davros

Not wishing to be rude, but it's considered polite to start your own thread in which to ask your questions, rather than hijacking a thread started by someone else for their questions :smile:

sorry relatively new to this so didnt know

thanks for the advice

Reply 12

TenOfThem

Original post by FranticR

Still there was no need for your over-reaction. I'm new, I'm doing my best, you've not given me a very good welcome.

And yet, I gave a lot of help and you have not even acknowledged it.

Why focus on the negative post instead of using the help I gave to move on?

Reply 13

davros

Original post by ilovechemistry2

sorry relatively new to this so didnt know

thanks for the advice

No problem!

Hopefully you can see the big orange (on my screen) button with Post new Thread (or something similar) written on it on the main forum page :smile:

Reply 14

FranticR

I have acknowledged it, I have been working on the questions since you posted.

Reply 15

ilovechemistry2

Original post by davros

No problem!

Hopefully you can see the big orange (on my screen) button with Post new Thread (or something similar) written on it on the main forum page :smile:

yeah got it now!

its under a new thread 'M1 Maths help'