Professor Hu did it here. Take his derivation for max range, and height, and show that Rmax=4H
www-hep.uta.edu/~yu/teaching/summer08-1441-001/lectures/phys1441-summer08-060308.ppt -
I failure to find an observation related to motion in two dimension projectile motion.
Prove That/Show that
maximum range =4Height
or
Rmax=4H
please help me thanks Ahsan Khan
4 answers
That link did not work
vertical v = Vo - gt
at the top, H, v = 0
so
t = Vo/g
then H = Vo t - .5 g t^2
H = Vo^2/g -.5 Vo^2/g = .5 Vo^2/g
that t is time to the top, H, which is half the total time in the air
u = Uo constant horizontal speed
R = range = Uo *2t = 2 Uo t = 2 UoVo/g
so
H/R = .5Vo/2 Uo = (1/4) Vo/Uo
I suspect that you class has already done the angle for maximum range is 45 degrees above horizontal where Vo = Uo
so
H/R = 1/4
vertical v = Vo - gt
at the top, H, v = 0
so
t = Vo/g
then H = Vo t - .5 g t^2
H = Vo^2/g -.5 Vo^2/g = .5 Vo^2/g
that t is time to the top, H, which is half the total time in the air
u = Uo constant horizontal speed
R = range = Uo *2t = 2 Uo t = 2 UoVo/g
so
H/R = .5Vo/2 Uo = (1/4) Vo/Uo
I suspect that you class has already done the angle for maximum range is 45 degrees above horizontal where Vo = Uo
so
H/R = 1/4
Anyway in case you have not done it, for speed S and elevation angle T:
Uo = s cos T
Vo = s sin T
for max height
t = (s/g) sin T like before
R = 2 Uo t = 2 s cos T (s/g )sin T
for max
dR/dT = 0
(2 s^2/g)(-sin^2 T + cos^2 T)
or max range when T = 45 degrees and
Uo = Vo
Uo = s cos T
Vo = s sin T
for max height
t = (s/g) sin T like before
R = 2 Uo t = 2 s cos T (s/g )sin T
for max
dR/dT = 0
(2 s^2/g)(-sin^2 T + cos^2 T)
or max range when T = 45 degrees and
Uo = Vo
i don't know the answer but how 4 the man height varies
1 answer