Help with Projectile Question

  • Thread starter Nate-2016
  • Start date
  • #1
9
0

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, im 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
 

Answers and Replies

  • #2
Having trouble understanding your notation (0 instead of [itex] \theta [/itex]) but one mistake I'm seeing right away is the following: in the equation that you're applying, the [itex] s [/itex] 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?
 
  • #3
9
0
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. Im 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.
 
  • #4
9
0
84 sin^2θ = sqrt (2 x (-9.81) x 40 )
sin-1(28.01/84)
= 19.5 degrees

Thank you
 
  • #5
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: [tex] x=x_0+v_0t+\frac{1}{2}at^2 \\ v=v_0+at [/tex]
Try getting the formula you used from these other two by eliminating the time, it's a worthwhile exercise ;)
 

Related Threads on Help with Projectile Question

Replies
2
Views
2K
Replies
3
Views
4K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
4
Views
914
Replies
4
Views
1K
Replies
7
Views
2K
Replies
5
Views
1K
Replies
3
Views
6K
Replies
0
Views
819
Top