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Projectile Motion Problems- Find Angle

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    (e) What is the angle made by the velocity vector with the horizontal?

    A projectile is shot from the edge of a cliff h = 165 m above ground level with an initial speed of v0 = 145 m/s at an angle of 37.0° with the horizontal, as shown below:

    Seconds in the air is 19.532760239244074174748994000901s
    Range of total distance traveled is 2261.9356107635250879761691838034m
    Final Velocity for x before it hits the ground: 115.80214895685746271118011659m/s
    Final Velocity for y before it hits the ground: -104.15787198754492632443347614472m/s (upwards is positive)
    Magnitude of final velocities (hypotenuse-more or less): 155.75300960174092250733936761367m/s

    These are just other informations obtained throughout the problem.

    2. Relevant equations

    3. The attempt at a solution

    Basically I used opposite divided by adjacent and then inverse tan it.
    Tan θ = opp/adj
    Tan θ = 165m/2261.9356107635250879761691838034m
    θ = 4.172129598188136156302219438937°
    Here is a picture of it, very slightly modified:
    [img=http://img583.imageshack.us/img583/733/projectilelaunchangleik5.th.jpg] (the picture wont show up as image)

    This did not come out as correct. If anyone can point out errors to find the correct answer, it would be appreciated.
  2. jcsd
  3. Nov 24, 2008 #2


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    Homework Helper

    All those decimal places make it quite hard to follow (just a tip, if you want very precise answers, use the exact answer rather than a string of decimals) but the only mistake you made that I can quickly spot out is that you used

    tan[tex]\theta[/tex]=(vertical displacement) / (horizontal displacement)

    rather you need to use the tangent for the horizontal and vertical velocity vectors to find the angle it makes with the horizontal.
  4. Nov 24, 2008 #3
    Thanks. That was correct. I see what I did wrong. Thank you.
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