Physics Projectile motion

Original post by Hasham123

A particle is projected vertically upwards at 20ms from point P.1s later a second particle is projected vertically upwards at same velocity from P . How long after second particle was projected do the 2 particles collide
Answer is 1.54 from start of second one
How to get 1.54

The balls collide when they have the same displacement from the starting point.

You can use one of the suvat equations (which includes s, u, t, and a) on each of the particles. If particle 1 is at time

t

in its flight, then particle 2 will be at time

t-1

in its flight since it was launched one second later.

Set the displacements equal to each other and solve for t.

Please post your workings :smile:

Reply 2

OP

So do s= uvxt +1/2at^2 assign s a value say 5 then do
5=uvxt +1/2at^2
5=uvx(t-1) +1/2ax(t-1)2

Reply 3

Original post by Hasham123

So do s= uvxt +1/2at^2 assign s a value say 5 then do
5=uvxt +1/2at^2
5=uvx(t-1) +1/2ax(t-1)2

That's basically the right method, but check your SUVAT equations... should be

s=ut+\frac{1}{2}at^2

, where u=20m/s and a = g = -9.81m/s^2.

Why are you setting the value of

s=5

?

You have two equations with s=..., therefore both equations equal each other.

Hence you can solve for t.

(edited 6 years ago)

Reply 4

the bear

20

ut + 0.5*-9.8*t² = u(t-1) + 0.5*-9.8*(t - 1)²

no need to put in a number such as 5.

put u = 20 m/s into both sides.

OP

So S=20t -4.9t^2

OP

Wait is it simaltanius equations

Reply 7

Original post by Hasham123

Wait is it simaltanius equations

Yes, you have two equations with s=... therefore you can solve them simultaneously by setting them equal to each other so there are just ts.

Reply 8

OP

So
S=20t-4.9t^2
S=20(t-1) -4.9(t-1)^2
Multiply out the brackets then solve for t?

Reply 9

Original post by Hasham123

So
S=20t-4.9t^2
S=20(t-1) -4.9(t-1)^2
Multiply out the brackets then solve for t?

Yes, so

20t-4.905t^2 = 20(t-1)-4.905(t-1)^2

And solve as you said.

Note that in AS physics, most boards ask for

g=9.81

, whereas maths exams tend to use

g=9.8

Reply 10

OP

Amateur question but do I times the bracket by outside number before squaring it or square it then times the answer by outside number

Reply 11

Original post by Hasham123

Amateur question but do I times the bracket by outside number before squaring it or square it then times the answer by outside number

Multiply out the brackets using FOIL, then multiply each term by the constant.

Reply 12

OP

So do (t-1)^2 then times answer by -4.905

Reply 13

Original post by Hasham123

So do (t-1)^2 then times answer by -4.905

Yes, but you need to expand

(t-1)^2

using FOIL:

Spoiler

OP

Foil?

Original post by Hasham123

Foil?

First, Outer, Inner, Last

It's a method for expanding brackets that you should have met at GCSE.

Note that

(t-1)^2 = (t-1)(t-1)

, so there you have a pair of brackets to expand.

Reply 16

OP

24.905=-49.5t
T=0.5031 which is not right

Reply 17

OP

Whoops that's wrong messed up

Reply 18

OP

24.905=9.8t
T=2.54
Then I just take away 1 then I'm at 1.54

Reply 19