projectile .. range formula Derivation?

Question

how can u get to .. R = [Vi^2(sin2θ)]/g ?

Answer 1

Start from the equations of motion
x(t)=Vi*cos(θ)*t
and
y(t)=Vi*sin(θ)*t-.5*g*t^2

when y(t)=0, the projectile is launched, and impacts at the range (it's a quadratic)

so
R=Vi*cos(θ)*t
find t when y(t)=0
0=Vi*sin(θ)*t-.5*g*t^2
we also know that
vy(t)=Vi*sin(θ)-g*t
when vy(t)=0, the projectile reaches apogee
therefore t at apogee is
Vi*sin(θ)/g
this is 1/2 the total flight time.
To prove,
plug
2*Vi*sin(θ)/g into the equation for y(t)

y(tR)=Vi*sin(θ)*2*Vi*sin(θ)/g-
.5*g*4*Vi^2*sin^2(θ)/g^2


simplify

y(tR)=2*Vi^2*sin^2(θ)/g-
2*Vi^2*sin^2(θ)/g

y(tR)=0, that shows that the projectile makes impact at twice the time to apogee.

so, now plug
2*Vi*sin(θ)/g

into x(t), which is expressed as R

R=Vi*cos(θ)*2*Vi*sin(θ)/g

simplify
R=Vi^2*cos(θ)*2*sin(θ)/g

Now use the identity
2*cos(θ)*sin(θ)=sin(2*θ)


R=Vi^2*sin(2*θ)/g

That's how you derive it

j

Answer 2

The "Range" formula for an object starting at (0,0) with velocity "vi" at angle "()" and returning at location, (x,0) is found by starting with the position eqs; x = viCos()t y = viSin()t - (1/2)gt^2 Solve the second eq, algebraically, for "t" when y=0. This gives the time of flight. Then substitute the expression found for "t" into the first eq. This gives the horizontal distance traveled in that time, and your desired eq.

Answer 3

first u need to find Time taken by projectile in total.
consider motion in vertical axis:
When body stops at highest point (v becomes 0)
v = u + at
t = (v - u) / a
t = 0 - usinQ / -g
t = usinQ/g
Now this is time taken to reach the top and stop.. so total time , by symmetry of motion, is equal to 2t
T = 2t
T = 2usinQ/g

Now for range:
consider motion along horizontal axis:
Range = speed * time
Range = ucosQ * 2usinQ/g
Range = u^2 . 2sinQ.cosQ/g
Range = u^2. sin2Q /g


CHEERS! :D