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Home Q & A Board Science-Related Homework Help Physics
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smont smont
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11 years ago
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.
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#1
11 years ago
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
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