M1 Projectiles

A particle is projected from a point O and passes through a point P when travelling horizontally. P is 10 m horizontally and 8 m vertically from O.

The angle of projection and the magnitude of the initial velocity need to be found but how do I begin to answer this question start?
Thank you!!!

Consider vertical and horizontal motion separately.

Note: It is travelling horizontally at P. Hence it's vertical velocity there is?
This will give you t in terms of initial velocity and the angle of projection.

Then standard s=ut+... horizontally and vertically.

Have a go.

I have used v=u+at and got 0=usin(theta) - 9.8t and rearranged to find t=usin(theta)/9.8 but where would I go from here?

As per my previous post, s=.... You'll get two equations in u and theta, to solve simultaneously.

As per my previous post, s=.... You'll get two equations in u and theta, to solve simultaneously.

From this I have formed two simultaneous equations as t=usin(theta)/9.8

8=u(usin(theta)/9.8)-4.9(usin(theta)/9.8)^2
10=u(usin(theta)/9.8)

is this right?

10=u(usin(theta)/9.8)cos(theta)

Hi, I am a bit confused as to how you would finish this off?

Post what you've got so far.

I have the 2 simultanoues equations: 8=u(usin(theta)/9.8)-4.9(usin(theta)/9.8)^2
and 10=u(usin(theta)/9.8)cos(theta) but not sure how to get an angle of projection from this

Those are just the equations as posted by others on here. You should be able to simplify the first one considerably, then divide one by the other and that will give you tan theta.

For the first equation I got 156.8= U^2sin thetha(2-sin thetha). Is this correct?

Actually, I just realised that original equation "8=u(usin(theta)/9.8)-4.9(usin(theta)/9.8)^2" is incorrect.

So, how did you get to that point?

Oh, I used s= vt- 0.5at^2 and subbed in the value for the height vertically

I'm puzzled as to how you arrived at the same, incorrect equation as previously posted on this thread if you used that formula.

Note, v=0 for the vertical motion so your first term will be zero.

If you've substituted correctly, you should now be able to do the division suggested in post #12

I actually realised my mistake and have got 8= -.0.5 *9.8* (Usin thetha/9.8)^2

Since the "8" implies you took upwards as postive, the acceleration would be -9.8.
Then carry on.