Recent content by Vikingjl11

Ok, knowing that makes this problem much easier. If the coordinates are 15,0,10, than the total distance in the x direction is 18.03m

To find the initial velocity we must be given a time interval or launch angle.

Is distance in ft. or meters? Also, there must be a time interval or a given angle.

Yea, your explanation certainly makes sense. Thanks for the insight.

Algebraic principle... ex. 100 - 5/5 = 99 100(5) - 5/5(5) = 99(5) correct 100 - 5/5(5) = 99(5) Incorrect

You must also multiply the bottom term by -V2z.(fourth step)

hootenanny- is the center of mass not moving at (V/2) m/s?

Yea, that would be equal to the x-component of L(lift) and -R.

There is no acceleration in either the x or y direction. This means the net force must be zero in both the x and y directions. Therefore, the upwards force must be equivalent to the magnitude(absolute value) of the downforce or weight(weight = M*g). Also, R must be equal to the magnitude of the...

Velocity is not a force. R is related to the horizontal component of the Lift. Your Trig. was also incorrect.

A Newton is a quantity of Force, you are attempting to find mass, acceleration due to gravity is a constant

52100Ncos(69) is incorrect...

yes, F(force)=M(mass) x a(acceleration)- In this case "a" equals accel. due to gravity.

Is that 1.47 x 102 Newtons? F = ma(gravity)

Vertical component of Lift must be equivalent to the magnitude of the weight.