Recent content by Nate-2016

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    Projectile Motion: Solving for Angle and Time

    84 sin^2θ = sqrt (2 x (-9.81) x 40 ) sin-1(28.01/84) = 19.5 degrees Thank you
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    Projectile Motion: Solving for Angle and Time

    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...
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    Projectile Motion: Solving for Angle and Time

    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...
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    Projectile Angle and Distance problem

    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) = (?)
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    Projectile Angle and Distance problem

    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
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    Projectile Angle and Distance problem

    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.
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    Projectile Angle and Distance problem

    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...
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    Help with variable mass on a slope question

    Thank you for explaining that to me. I have now taken into consideration the angle of the slope at the very end of my equation. My new equation is now 0 = mdv - Mv dv+v = 650m/s mdv = Mv mx(650-v) = Mv 15x(650-v) = 5000v v = (15x650) / (15+5000) v = 1.94 m/s x cos(tan^1(5/12)) v = 1.79 or 1.8...
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    Help with variable mass on a slope question

    Homework Statement An armoured car with a mass of 5 tonnes is located on a smooth plane which is inclined at an angle of tan-1 (5/12) to the horizontal as shown in Figure Q3. A missile of mass 15kg is fired horizontally from this armoured car at 650m/s. Determine the velocity with which the...
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