projectile motion, how to find initial velocity

[sweer6]

I don't understand how to calculate the initial velocity, from the information given below.

The initial height is 1.95m (the intial horizontal displacement is 0), the angle of release is 35 degrees and the range of the projectile (or the horizontal displacement at impact) is 90.33m. How do i find the initial velocity?

Some one else told me to use this equation and solve for v,
but then i need to know how the equation below is derived from the basic displacement, velocity and acceleration equations/ vectors. I need to know how this formula is created from the basics up, like, i know that a = -g j and that v = v -gt + c1 j and that r = 0.5gt^2+c1 +c2, and that c1 in velocity vector = v(initial) cos theta (angle) + v(initial) sin theta (angle) and so on, i hope i din't make any mistakes here but what i want to know is how the above equation in the image is derived from the ones that i just mentioned. Sorry if something is wrong or i have forgotten

 

[HallsofIvy]

I don't understand how to calculate the initial velocity, from the information given below.

The initial height is 1.95m (the intial horizontal displacement is 0), the angle of release is 35 degrees and the range of the projectile (or the horizontal displacement at impact) is 90.33m. How do i find the initial velocity?

Some one else told me to use this equation and solve for v, View attachment 4137
but then i need to know how the equation below is derived from the basic displacement, velocity and acceleration equations/ vectors. I need to know how this formula is created from the basics up, like, i know that a = -g j and that v = v -gt + c1 j and that r = 0.5gt^2+c1 +c2, and that c1 in velocity vector = v(initial) cos theta (angle) + v(initial) sin theta (angle) and so on, i hope i din't make any mistakes here but what i want to know is how the above equation in the image is derived from the ones that i just mentioned. Sorry if something is wrong or i have forgotten

It's probably simplest to work with the horizontal and vertical component separately. The vertical acceleration is -g and the horizontal acceleration is 0. If the initial speed is "v" then the horizontal component of speed is v cos(35) and the vertical component of speed is v sin(35)- gt. With the initial position as (0,1.95), the horizontal position is x= v cos(35)t and y= 1.95+ v sin(35)t- (g/2)t^2. Knowing that the range is 90.33m means that, at some time, t, x= v cos(35)t= 90.33 and y=1.95+ v sin(35)t- (g/2)t^2= 0. That is two equations for v and t.

From the first equation, t= 90.33/(v cos(35)). You can replace the t in the second equation with that getting a single equation to solve for v.

 

[wjm11]

I don't understand how to calculate the initial velocity, from the information given below.

The initial height is 1.95m (the intial horizontal displacement is 0), the angle of release is 35 degrees and the range of the projectile (or the horizontal displacement at impact) is 90.33m. How do i find the initial velocity?

Some one else told me to use this equation and solve for v, View attachment 4137
but then i need to know how the equation below is derived from the basic displacement, velocity and acceleration equations/ vectors. I need to know how this formula is created from the basics up, like, i know that a = -g j and that v = v -gt + c1 j and that r = 0.5gt^2+c1 +c2, and that c1 in velocity vector = v(initial) cos theta (angle) + v(initial) sin theta (angle) and so on, i hope i din't make any mistakes here but what i want to know is how the above equation in the image is derived from the ones that i just mentioned. Sorry if something is wrong or i have forgotten

You might find this useful if you're interested in the derivation(s):

http://en.wikipedia.org/wiki/Range_of_a_projectile

 

[kaamrev1]

If the projectile's position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for v0 in the aforementioned parabolic equation:

.