Mechanics M1- Projectiles (please help)

hi, I have been trying to do this question and I can't figure out where I'm wrong. here's the question:

A ball is thrown from a height of 1.5m at a speed of 12ms^-1 and at and angle of 20 degrees above the horizontal. it hits the ground for the first time T seconds after it was thrown.
(a) Find T
(b) Find the horizontal distance travelled by the ball.
(c) Find the speed of the ball when it hits the ground.

Now, for part a I used y=utsintheta - 1/2gt^2 and from that I got a quadratic equation which was 4.9t^2 -12tsin20 -15 =0 and by solving this I got T=2.21786 and T=-1.380 but surely time can't be negative so it must be T=2.21786 but the answer is 1.11s somehow! please help!

Original post by Sid.

Now, for part a I used y=utsintheta - 1/2gt^2 and from that I got a quadratic equation which was 4.9t^2 -12tsin20 -15 =0 and by solving this I got T=2.21786 and T=-1.380 but surely time can't be negative so it must be T=2.21786 but the answer is 1.11s somehow! please help!

What's your thought process on getting the -15 within the quadratic??

Reply 2

Sid.

OP

9

Hi again, because it's thrown from a height so I used y=utsintheta -1/2gt^2 +h and not y=utsintheta-1/2gt^2 (sorry) so using the formula I got 0= 12tsin20 -4.9t^2 +15 so making the 4.9 positive gives -15

Original post by Sid.

Hi again, because it's thrown from a height so I used y=utsintheta -1/2gt^2 +h and not y=utsintheta-1/2gt^2 (sorry) so using the formula I got 0= 12tsin20 -4.9t^2 +15 so making the 4.9 positive gives -15

OK, do you realise that it should be 1.5 and NOT 15? (REREAD THE QUESTION)

Reply 4

Muttley79

20

Original post by Sid.

Hi again, because it's thrown from a height so I used y=utsintheta -1/2gt^2 +h and not y=utsintheta-1/2gt^2 (sorry) so using the formula I got 0= 12tsin20 -4.9t^2 +15 so making the 4.9 positive gives -15

It lands below the point where it was thrown.

Set out your variables:
s = -1.5 [if upwards is positive and you are defining zero as the starting point]
a =
u = [vertical component]

Reply 5

Sid.

OP

9

Original post by RDKGames

OK, do you realise that it should be 1.5 and NOT 15? (REREAD THE QUESTION)

oh dear, thanks pal! It works Now! I will let you know if I need any more assistance on this question😊

Reply 6

Sid.

OP

9

Original post by RDKGames

OK, do you realise that it should be 1.5 and NOT 15? (REREAD THE QUESTION)

also, any hints on part c? I'm thinking SUVAT BUT I'M UNSURE WHETHER S=1.5 OR 0?

Original post by Sid.

also, any hints on part c? I'm thinking SUVAT BUT I'M UNSURE WHETHER S=1.5 OR 0?

t=1.11

when it hits the ground. So determine the velocity vector at that time (meaning you need to get the horizontal and vertical velocities and construct a vector containing them). From there you can calculate the speed.

Reply 8

Sid.

OP

9

Original post by RDKGames

t=1.11

when it hits the ground. So determine the velocity vector at that time (meaning you need to get the horizontal and vertical velocities and construct a vector containing them). From there you can calculate the speed.

why can't I use SUVAT instead?

Original post by Sid.

why can't I use SUVAT instead?

You can... that's what you're using to determine the individual velocities.

Reply 10

JARD112358

6

To work out the time you need to use suvats. This is because you need work vertically rather than horizontally and vertical motion is calculated by suvats. Hope this helps.

Reply 11

Sid.

OP

9

Original post by RDKGames

You can... that's what you're using to determine the individual velocities.

I have got the answer by finding each velocities. for horizontal I used: ucostheta and vertical I used: utsintheta-gt which gives me the correct answer so I didn't use SUVAT but now you saying you can. I'm confused now even though I have done the question!

(edited 6 years ago)

Original post by Sid.

I have got the answer by finding each velocities. for horizontal I used: ucostheta and vertical I used: utsintheta-gt which gives me the correct answer so I don't use SUVAT but now you saying you can. I'm confused now even though I have done the question!

You used SUVAT to work out the horizontal and vertical velocities individually which is what I said!

Reply 13

Sid.

OP

9

Original post by RDKGames

You used SUVAT to work out the horizontal and vertical velocities individually which is what I said!

I meant can't you use SUVAT SO S=1.5, U=12, V=?, A=? AND T=1.11 SO FROM THAT I CAN WORK OUT V RIGHT?

Original post by Sid.

I meant can't you use SUVAT SO S=1.5, U=12, V=?, A=? AND T=1.11 SO FROM THAT I CAN WORK OUT V RIGHT?

Yes I understood what you said (you're repeating yourself), and I replied in post 10 saying that's exactly what you need to use and do...

What's the confusion??

Reply 15

Sid.

OP

9

Original post by RDKGames

Yes I understood what you said (you're repeating yourself), and I replied in post 10 saying that's exactly what you need to use and do...

What's the confusion??

let me explain, so I tried using S=(U+V)T/2 SO rearranging that gives me 2S/T - U= V? And solving this gives me a negative answer of -9.29 which is surely not the answer

Original post by Sid.

let me explain, so I tried using S=(U+V)T/2 SO rearranging that gives me 2S/T - U= V? And solving this gives me a negative answer of -9.29 which is surely not the answer

Is this for horizontal or vertical velocity?

Also what numbers did you use?

Velocities can very well be negative.

Mechanics M1- Projectiles (please help)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you