The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
A is the answerrrrrrrrrrrr
I beleive that the answer is B.
Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
Mechanical and chemical. Mechanical breaks down food into smaller pieces. Chemical breaks it down into simpler nutrients that can then be used by cells. Hope this helps!
