Projectile Angle and Distance problem

[Nate-2016]

Homework Statement

A projectile was launched from the top of a 100m tower upward at an angle theta to the horizontal with an initial velocity of 36m/s. The time of flight of the projectile was 8 seconds.

Determine:
i) The angle of projection
ii) The maximum height reached by the projectile

Homework Equations

S=ut+1/2at^2

The Attempt at a Solution

I am used to solving these questions when i am not given time but given both the horizontal and vertical distances. At which I use the quadratic formula to solve for theta. However in this case I can not do that as when i split my horizontal and vertical components and apply the above equations I am left with these.

Horizontally
S=36cos(?)*8

Vertically
-100=36sin(?)*8+0.5*-9.81*64

I have tried rearranging the vertical formulas to solve for theta

-1/2*-9.81*64-100/8=36sin(?)
301.42 = 36*sin(?)
sin-1(36/301.42) = 6.85 degrees

I really don't think that the answer is correct. The lecture who set the question will not give any answers.

 

[TSny]

Vertically
-100=36sin(?)*8+0.5*-9.81*64

OK

I have tried rearranging the vertical formulas to solve for theta

-1/2*-9.81*64-100/8=36sin(?)

You have only the 100 being divided by 8. But is that correct?

 

[Nate-2016]

Yeah. Only the 100 divided by 8. I am thinking that maybe 6.85 degrees is the correct angle for a projectile to fall from 100 metres in 8 seconds.

 

[TSny]

Only the 100 divided by 8.

Check the algebra on this. I don't think it's correct to have just the 100 divided by 8.

 

[Nate-2016]

Im not exactly sure on the rerrangment to solve for U. As i thought my 1st rearrange was correct.

I tried the rearrange dividing the whole thing by 8 = sin-1(26.74/36) = 47.96 degrees.

I tried the rearrange dividing by 16 from 1/2*8 under everything = sin-1(36/60.76) = 36.33 degrees

 

[Nate-2016]

s = ut + 1/2at^2
ut = s - 1/2at^2
u = (s - 1/2at^2)/t
36sin(?) = (-100-1/2*-9.81*64)/8

Sin(?)= 26.74/36
Sin-1(26.74/36) = (?)

 

[TSny]

s = ut + 1/2at^2
ut = s - 1/2at^2
u = (s - 1/2at^2)/t
36sin(?) = (-100-1/2*-9.81*64)/8

Sin(?)= 26.74/36
Sin-1(26.74/36) = (?)

That looks correct.