Projectile Motion: Solving for Angle and Time

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A projectile is launched from a 200m tower at an initial velocity of 84m/s, reaching a maximum height of 240m. The angle of projection is calculated to be 19.5 degrees, with a flight time of 4.14 seconds. The discussion highlights the importance of correctly identifying displacement in projectile motion equations, clarifying that the relevant displacement is the height above the tower, which is 40m. Participants emphasize the need to derive equations from fundamental motion principles for better understanding. The solution ultimately confirms the correct angle and time of flight for the projectile.
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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
 
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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?
 
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.
 
84 sin^2θ = sqrt (2 x (-9.81) x 40 )
sin-1(28.01/84)
= 19.5 degrees

Thank you
 
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 ;)
 
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