A projectile is launched at a certain speed at an angle of 27.5 ...

[camila]

A projectile is launched at a certain speed at an angle of 27.5 degrees from the horizontal.  Find the ratio of maximum height to the range.  That is find ymax/R. Hint:  divide the equation for ymax by the equation for R.

[Jones]

If the launch speed is V, the horizontal component of velocity is
V.cos(27.5)
and the vertical component is
V.sin(27.5)
At max height v = 0
v[y] = u[y] + a[y].t
t = (v[y] - u[y])/a[y] = (0 - V.sin(27.5))/(-9.8 ) = V.sin(27)/9.8
Height reached
v[y]^2 - u[y]^2 = 2as
H = (v[y]^2 - u[y]^2)/2a = [Vsin(27.5)]^2/(2 x 9.8 )
Time of flight = 2V.sin(27.5)/9.8
range
R = u

• .t = [V.cos(27.5).2V.sin(27.5)]/9.8

H/R = sin(27.5)/cos(27.5) = 0.52

[camila]

Thanks! It was right