Derive Equation for Launch Angle | Physics Solved Variables

[Remulak]

How do you derive the equation for the launch angle using the range and peak equations? This isn't a homework problem but something i want to know for my own general knowledge for solving physics problems. Is there a site that has physics equations solved for different variables? Hyperphysics is a great site but it doesn't work through the deriving and solving of the equations.

 

[bel]

Given its height, you can find the upward component of its initial velocity, and you can also find the time it is in flight, since you know the constant acceleration g. Given its range, and having found the time interval for which the projectile was in flight, you can find the horizontal component of its velocity. Having found those two componemts, the tangent of the angle is the ratio of the vertical component by the horizontal component of the velocity.

 

[Remulak]

how so if you have the y = yo + voyt - 1/2gt^2?, you'll have two unknown voy and t

 

[asd1249jf]

voy is your initial velocity. The problem MUST have some indication of velocity on the y-axis otherwise, it would be unsolvable.

 

[Remulak]

A problem that I'm working on has a range of 25m and a height or peak of 4.90m. It says find the initial velocity,the angle at which the projectile is fired, and the time its in the air. I looked on hyperphysics and it said if range and peak are given an equation can be derived to solve for the angle.

 

[asd1249jf]

A problem that I'm working on has a range of 25m and a height or peak of 4.90m. It says find the initial velocity,the angle at which the projectile is fired, and the time its in the air. I looked on hyperphysics and it said if range and peak are given an equation can be derived to solve for the angle.

Ooo OK.

We know the total traveled distance is 25m

And the max height is 4.9m

The equation for the velocities are

[tex]v_x = v_0cos(\theta)[/tex]
[tex]v_y = v_0sin(\theta)-gt[/tex]

And for distances are

[tex]s_x = v_0cos(\theta)t[/tex]
[tex]s_y = v_0sin(\theta)t-\frac{1}{2}gt^2[/tex]

First of all, plug 25m into the [tex]s_x[/tex] equation, solve for t to find the expression for
the total time of travel.

Then, plug in t into [tex]s_y[/tex] you still have [tex]v_0[/tex], [tex]theta[/tex] left as unknowns. Apparently, the y distance has to equal to 0 at this particular time you have found. Solve for [tex]\theta[/tex]. Now you have the derived expression for [tex]\theta[/tex].

To find the actual values, you need the last piece of information, where the max height is 4.9m. It takes half of the time of total distance travel for the projectile to reach this max height. So plug in 4.9m as [tex]s_y[/tex], plug in t/2 into the expression, then it's just simultaneous equation - 2 equations and 2 unknowns

 

[learningphysics]

l46kok's explanation is excellent. Only thing I'd recommend is not plugging in actual numbers until the very end... meaning get the formula for arbitrary range R, and height H...

Then once you have the equations for velocity and angle... then plug in the actual numbers...

 

[Remulak]

thanks for the explanation