Posted by **Muhammad Ahsan Khan** on Thursday, January 1, 2009 at 12:06pm.

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

- Physics -
**bobpursley**, Thursday, January 1, 2009 at 4:41pm
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 -

- Physics -
**Damon**, Friday, January 2, 2009 at 8:49am
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

- Physics - additional -
**Damon**, Friday, January 2, 2009 at 8:57am
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

- Physics -
**karan **, Friday, February 13, 2015 at 3:51am
i don't know the answer but how 4 the man height varies

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