Projectile motion absolutely puzzled

[robcowlam]

I am trying to calculate the initial velocity, u and angle of projection(theta) of a projectile. The projectile is fired from 1.06m above the ground and from this point reaches a maximum of 2.12m from the ground. The distance traveled in the x-direction when the projectile reaches its maximum height is 1m.
This is all the information given in the question.

I am taking my x-axis to be the horizontal direction and the y-axis the vertical direction, calling the point where the projectile is fired from the origin.

I have tried to get a solution in the following way:
Motion in y-direction:
a=dv/dt
a dt = dv
integrating both sides gives:
a.t = 0 - uSin(theta) (where u is initial velocity and v is final velocity)
dividing by a gives an expression for the time taken to reach maximum height.

Motion in the x-direction:
S=ut +0.5at^2
however a = 0
so; 1=uCos(theta).t

Substituting in t from earlier gives:
9.81=(u^2) Cos(theta)Sin(theta) from which i get u^2=9.81/sin(theta)cos(theta)

In the y direction again:
v^2 -u^2 = 2as
20.8=9.81/sin(theta)cos(theta)
9.81/20.8=sin(theta)cos(theta)
2(0.47)=2sin(theta)cos(theta) using the trig ID sin(2A)=SinACosA
sin(2theta)=0.94
and theta =35degrees;

Any help would be GREATLY appreciated
Thanks!


However this is not the answer given with the problem

 

[Dadface]

Hello rob,If I'm following you correctly you seem to have three unknowns(u,t and theta) but two equations.

For motion in the y direction and to reach the greatest height you have three givens

Final velocity at top(v)=0
distance(s)=1.06(you didnt seem to use s in your attempt)
acceleration(a)=g
With this you could use a suitable equation of motion eg
v squared = u squared -2gs

 

[robcowlam]

ok I see what you're saying, I've had another go and I've got 3 equations and 3 unknowns but I can't seem to solve them simultaneously.
My 3 equations are:
2=U.t.Cos(theta)
1.06=U.t.Sin(theta)-4.905t^2
0=USin(theta)-9.81t

Ive tried to rearrange each equation and substitute them into each other but it gets really complicated and I don't seem to get the right answer. I thought of using a matrix and gauss' elimination to do it but am unsure how to build up an augmented matrix as the variables will not separate.
Any advice?

 

[Dadface]

Hello rob.Oh the deep joy of getting bogged down in simultaneous equations.
From your equation three... usin(theta)= 9.81t
subbing into equation two we get:
1.06=9.81t^2-4.905t^2=4.905t^2.From this t can be found
Look again at your equation one above.When the projectile reaches its greatest height the horizontal distance traveled is one metre not two.

usin(theta)=9.81t...(a) and
ucos(theta)=1/t...(b)
sin(theta)/cos(theta)=tan(theta) so if (a) is divided by(b) the u cancels and we have an expression for tan(theta) that can be subbed back to find u.

 

[robcowlam]

Got it! I was overcomplicating it by obtaining an expression for t instead of usin(theta)!
Thanks for your help!