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...
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...
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
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...
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...
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...