Question

the maximum height and horizontal range are equal for a projectile launched upward from ground at an angle horizontal equal to

Answer 1

this will help here too..
R/H = 4U/V

Answer 2

options are 60 90 tan^-1/8 tan^-1/4

Answer 3

plzz help

Answer 4

put R=H in above equation
U= u cos(theta)
V= usin(theta)
Find theta..

Answer 5

i tried but not getting plzzz do this plzz

Answer 6

substituting values in above equation
\[\theta =\tan^{-1}(4)\] Didn't you get it?

Answer 7

no plzz can u go it by step by step

Answer 8

Put R=H, that makes R/H=1
Now R/H= 4U/V
or, 1=4 (u cos(theta)) /(u sin (theta))
that makes, tan (theta) = 4

Answer 9

hey R=u^2sin2theta/g u have written cos???

Answer 10

XX' : x=Vo*cosθ*t
YY': y=Vo*sinθ*t-1/2*g*t^2
set y=0 and solve to t to find total time of flight
put that to x=..... to find x max.

YY' Vy=Vo*sinθ-g*t
put Vy=0 and solve to t to find time for max height
put that t to y=Vo*sinθ*t-1/2*g*t^2 to find h max
mast be Xmax=Hmax
substitude and see what you will get....

Answer 11

can some one make it clear

Answer 12

oh.........no

Answer 13

can we use R=u^2sin2theta/g H=u^2sin^2theta/2g

Answer 14

ok

Answer 15

Since R=H
u^2sin2theta/g =u^2sin^2theta/2g
or, 2sin2theta= sin^2 theta
or 2*2 sin theta * cos theta = sin^2 theta
or, tan theta =4

Do you get it now?