Answer:
We are given the trajectory of a projectile:
y=H+xtan(θ)−g2u2x2(1+tan2(θ)),
where H is the initial height, g is the (positive) gravitational constant and u is the initial speed. Since we are looking for the maximum range we set y=0 (i.e. the projectile is on the ground). If we let L=u2/g, then
H+xtan(θ)−12Lx2(1+tan2(θ))=0
Differentiate both sides with respect to θ.
dxdθtan(θ)+xsec2(θ)−[1Lxdxdθ(1+tan2(θ))+12Lx2(2tan(θ)sec2(θ))]=0
Solving for dxdθ yields
dxdθ=xsec2(θ)[xLtan(θ)−1]tan(θ)−xL(1+tan2(θ))
This derivative is 0 when tan(θ)=Lx and hence this corresponds to a critical number θ for the range of the projectile. We should now show that the x value it corresponds to is a maximum, but I'll just assume that's the case. It pretty obvious in the setting of the problem. Finally, we replace tan(θ) with Lx in the second equation from the top and solve for x.
H+L−12Lx2−L2=0.
This leads immediately to x=L2+2LH−−−−−−−−√. The angle θ can now be found easily.
That could be a symptom but I recommend going to your doctor just to be on the safe side!
Answer:
appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity
Explanation:
Answer:
Kepler's Third Law
T = 2 π r 3 G M E . T = 2 π r 3 G M E . For an ellipse, recall that the semi-major axis is one-half the sum of the perihelion and the aphelion. For a circular orbit, the semi-major axis (a) is the same as the radius for the orbit.
volume of milk is given as
we will convert it into mL unit
mass of the milk m = 2kg
m = 2000 g
now for the density we can use
<em>so the density is 0.264 g/mL for above sample</em>
