Projectile Motion: Solving for Angle and Time

[Nate-2016]

Homework Statement

A projectile is launched from the top of a 200m tower upwards at a non-vertical angle above level horizontal ground with an initial velocity of 84m/s and rises to a maximum height of 240m above ground level. Determine the angle of projection and time and flights.

Answer is (19.5 degrees, 4.14 seconds)

Homework Equations

I know i have to use the V^2 = U^2 + 2as and set V^2 to zero. But because the U is squared and I have 84sin0, I am unsure on how to solve for theta.

The Attempt at a Solution

I have 2 attempts

0=84sin0+2*(-9.81)*240
-2/-9.81x 240 = 84sin0^2
sqrt (8.49 x 10-4) = 84sin0
sin-1(0.029/84) = 0
0 = 0.197

84sin0^2=sqrt(2-9.81x240)
sin0^2=68.62/84
0=54.77

 

[guitarphysics]

Having trouble understanding your notation (0 instead of \theta) but one mistake I'm seeing right away is the following: in the equation that you're applying, the s stands for displacement. So if the projectile is launched from the top of the 200m tower, and reaches a max. height of 240m; is its displacement really 240m?

 

[Nate-2016]

yeah i replaced theta with 0.

Im not sure, I have used this question to solve for maximum height of the projectile, so I was under the assumption I could use this equation backwards. I am not sure what the actual displacement is. I think that when i do it usually i only take into consideration the height reached above the tower so i shall substitute that in and try again and let you know how i get on. Thank you for your help.

 

[Nate-2016]

84 sin^2θ = sqrt (2 x (-9.81) x 40 )
sin-1(28.01/84)
= 19.5 degrees

Thank you

 

[guitarphysics]

Nice! Glad you got it :)
For future reference: the equation refers to the displacement in the motion, since it comes out of the following two equations: x=x_0+v_0t+\frac{1}{2}at^2 \\ v=v_0+at
Try getting the formula you used from these other two by eliminating the time, it's a worthwhile exercise ;)