A projectile is launched from the ground at an initial velocity, v, at and angle of θ....?
. Determine the value of θ that would give the maximum range of the projectile. Also, determine the maximum range in terms of g and v.
It is not specified that this take place on Earth. Use g for acceleration due to gravity rather than substituting in a numeric value.
(use optimization equation if possible)
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Range in projectile motion is given by:
R = (v^2/g)sin2θ where v is the initial velocity vector in the direction of launch and θ is the launch angle.
See, http://www.pa.uky.edu/~moshe/phy231/lecture_notes/...
As it turns out (and you might have thought this intuitively) the maximum range occurs when the angle θ is 45 degrees.
It's easy to show this by differentiating the range formula with respect to theta.
R' = 2(v^2/g)cos(2θ)
setting R' equal to 0 and solving for θ, you'll find you get a local maximum at θ = pi/4 = 45 degrees.
From there, now that you know that 45 degress or pi/4 results in the maximum, it's easy to express the maximum range in terms of g anv v.
R = (v^2/g)sin2θ
Rmax = (v^2/g)sin(pi/2) = v^2/g
the vertical factor to velocity is 60 x sin30 =30 m/s at max ht this velocity is going to 0 substituting in v^2 = u^2 - 2gh 0 = 30 x 30 - 2 x 9.8 x H H = 30 x 30 / 2 x 9.8 = 46m