Let
y_0 =y0= initial height
R= R=range
v_0v0 = initial velocity
thetaθ= launching angle of the projectile.
A)
When the projectile is landed, y=0. And If the initial height is doubled, then
0 =2 y_0 +v_0sintheta*t - 1/2g t^20=2y0+v0sinθ⋅t−12gt2
rArrt =( v_0sintheta +-sqrt( (v_osintheta)^2+8gy_0))/g⇒t=v0sinθ±√(vosinθ)2+8gy0g
R= V_(0x)t=V_0costheta (( v_0sintheta +-sqrt( v_o^2sin^2theta+8gy_0))/g ) R=V0xt=V0cosθ⎛⎜
⎜⎝v0sinθ±√v2osin2θ+8gy0g⎞⎟
⎟⎠
Technically, you can solve for y_0y0 if R is known from the equation above, thus answering your part vice versa part# of your question.
Or you can do the following.
B) If you double R instead, it means R is a given.
v_(0x) = (Deltax)/(Deltat) rArr v_0costheta = (2R)/t
t= (2R)/(v_0costheta)
Again when the projectile landed, y=0
0=y_0 + v_0sintheta*t-1/2g t^2
y_0= v_0sintheta((2R)/(v_0costheta))-1/2g((2R)/(v_0costheta))^2
y_0=2R(tantheta-(gR)/(v_0^2cos^2theta))
