which one is angle will give the longest range ? 30 degrees ? 45 degrees ? 60 degrees ?
2 answers
It depends on the initial height. However, if initial height = 0, then 45 degrees will give you the longest distance.
range= VcosTheta*time
timeinair:
hf=ho+VisinTheta*t-1/2 g t^2
or 1/2 gt^2-ViSinTheta*t+(hf-ho)=0
and you solve for t. Clearly, the hf-ho term at first glance seems to matter. Lets check it.
using the quadratic equation.
t= (ViSinTheta+-sqrt(Vi^2Sin^2Theta-2g(hf-ho))/g
taking the + sqrt solution, put that into the horizontal equation..
range=VicosTheta(ViSinTheta/g +sqrt(Vi^2sin^2theta-2(hf-ho)/g)
now,with the assistance of calculus, maximizing range (drange/dtheta =0
0=Vi^2 [sin^2theta/g - cos^2theta/g)+1/2 1/sqrt( ) *2sinthetacostheta)
and you solve for theta
and the solution is a lot of algebra, but what I want to point out, the factor (Hf-ho) is in that squareroot function in the denominator,so as Max points out, it matters the difference in height.
timeinair:
hf=ho+VisinTheta*t-1/2 g t^2
or 1/2 gt^2-ViSinTheta*t+(hf-ho)=0
and you solve for t. Clearly, the hf-ho term at first glance seems to matter. Lets check it.
using the quadratic equation.
t= (ViSinTheta+-sqrt(Vi^2Sin^2Theta-2g(hf-ho))/g
taking the + sqrt solution, put that into the horizontal equation..
range=VicosTheta(ViSinTheta/g +sqrt(Vi^2sin^2theta-2(hf-ho)/g)
now,with the assistance of calculus, maximizing range (drange/dtheta =0
0=Vi^2 [sin^2theta/g - cos^2theta/g)+1/2 1/sqrt( ) *2sinthetacostheta)
and you solve for theta
and the solution is a lot of algebra, but what I want to point out, the factor (Hf-ho) is in that squareroot function in the denominator,so as Max points out, it matters the difference in height.