need help with projectile motion (1 Viewer)

[jemsta]

if someone can help it wouldbe greatly appreciated
an object is projected vertically upqards with a speed of 49m/s. 2 seconds later another object is projected vertically upwards from the same spot at the same speed. find when and where the two objects meet

 

[Dreamerish*~]

Are the angles given?

So far I've worked out that the displacement equations are:

x₁ = 49tcosθ

y₁ = 49tsinθ - 5t²

Since the next projectile is basically exactly the same as the first one, only it's projected 2 seconds later:

x₂ = 49(t + 2)θ

y₂ = 49(t + 2)θ - 5(t + 2)²

When they meet, x₁ = x₂ and y₁ = y₂.

I just can't work it out. If their angles are the same and their initial velocity are the same, and they are subjected to the same forces, then they just don't meet at all.

 

[Antwan23q]

yeh, use the formulars
so for the first one, y=49tsin@-1/2gt^2
but for the second one,
y=Vsin@(t-2)-1/2g(t-2)^2

because it is projected 2 seconds after, the time has to be t-2. hope it helps

 

[Antwan23q]

t-2 people

 

[Dreamerish*~]

nar there are no angles involved....must have some other way

Hmm, well I'm thinking no way can the angles be the same when the velocities are the same.

I have a brilliant idea! Let's wait for someone else to answer it.

 

[Antwan23q]

its been projected vertically upwards so it has no horizontal motion

 

[lucifel]

vertically upwards = angle = 90, work it out from there on in

 

[jemsta]

nar there are no angles involved....must have some other way

 

[Dreamerish*~]

t-2 people

Isn't it (t + 2) for the second projectile?

Because it's at the same spot as the first projectile two seconds later, hence adding two seconds to t?

 

[Antwan23q]

y=49-1/2gt^2= V-1/2g(t-2)^2

 

[Antwan23q]

Isn't it (t + 2) for the second projectile?

Because it's at the same spot as the first projectile two seconds later, hence adding two seconds to t?

nah, set t=0, it will say it is positioned 2 seconds ahead then. so u have to callibrate it, so when it has passed 2 seconds, it will be at the start. if u get me

 

[Dreamerish*~]

y=49-1/2gt^2= V-1/2g(t-2)^2

I totally don't get that.

Could you write it in better notation?

 

[Antwan23q]

lol
how about i just write it on paper and scan it in

 

[Dreamerish*~]

Lol, I managed to get t = 1 which is totally impossible because at t = 1, the second projectile hasn't even been fired yet.

 

[Antwan23q]

how odd. must be some mistake

 

[SaHbEeWaH]

it's being fired vertically up so there are no angles involved

projectile 1: consider vertical motion

a = -10
v = -10t + c
at t = 0, v = 49, so c = 49
v = -10t + 49
y = -5t^2 + 49t

projectile 2: consider vertical motion

a = -10
v = -10t + c
at t = 2, v = 49
49 = -20 + c
c = 69
v = -10t + 69
y = -5t^2 + 69t + c
when t = 2, y = 0
0 = -20 + 138 + c
c = -118
y = -5t^2 + 69t - 118

-5t^2 + 69t - 118 = -5t^2 + 49t
29t = 118
t = 4.07 seconds

EDIT: i can't subtract

-5t^2 + 69t - 118 = -5t^2 + 49t
20t = 118
t = 5.9 seconds

 

[Ghost1788]

does any one know the answer....the right answer cause i got t=3.37 which isnt what ^^^ got so um yea...i did my calculations seperately

 

[jemsta]

did you assume that g=9.8 or -9.8? coz i put 9.8 and it works but if you put -9.8 u get a negative. i got the answer which was 6 seconds and meets at 117.6metres, but if you do get a negative answer do u assume that t=6 since t >0?

 

[SaHbEeWaH]

I assumed it -10m/s^2

If you assume g = 9.8 then in theory throwing the projectile up will never fall down and since the ball is always accelerating up at a faster rate than the second you shouldn't even get an answer...

Velocity is a vector quantity and since gravity pulls objects down into earth hence the - sign

If you get a negative value of t don't just assume you're right and take the absolute value since t > 0... you should'nt ever get a negative value for t...

 

[acmilan]

You can take g = +9.8, you just have to switch the signs for anything going upwards. Both ways work.