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a is acceleration, v is velocity, s is position
a = - g
v = u - gt (by integrating with respect to t and setting initial speed u as constant of integration)
s = ut -gt2/2 + s0 (by integrating again and setting initial position s0 as constant of integration, although object if usually projected from origin so s0 = 0)
a = 0
v = u (constant velocity as acceleration is zero)
s = ut + s0 (again s0 usually 0)
so for an object projected from the origin at speed u at an angle θ from the horizontal, the initial speed in the x direction is ucos(θ) and usin(θ) in the y direction
thus sx = utcos(θ) and sy = utsin(θ) - gt2/2
flight time: this is the value of t when sy returns to zero
0 = utsin(θ) - gt2/2
so either t = 0 (at launch) or usin(θ) - gt/2 = 0
=> flight time is t = 2usin(θ)/g
flight distance: this is the value of sx when sy = 0
when sy = 0, t = 2usin(θ)/g
so sx = utcos(θ) = 2u2sin(θ)cos(θ)/g = u2sin(2θ)/g (as sin(2θ) = 2sin(θ)cos(θ))
=> flight distance is x = u2sin(2θ)/g
this is when sy is at a stationary point, ie. when dsy/dt = 0. This is also when vy = 0.
vy = usin(θ) - gt = 0
=> t = usin(θ)/g
=> sy = utsin(θ) - gt2/2 = u2sin2(θ)/g - u2sin2(θ)/2g = u2sin2(θ)/2g
=> greatest height is h = u2sin2(θ)/2g
Integration by parts states that ∫uv' dx = uv - ∫u'v dx, where u & v are funcions of x and the notation u' means du/dx.
to do integration by parts given an integral ∫f(x) dx, it involves writing f(x) as f(x) = u(x)v'(x), and then following the formula by determining u' and v.
The whole point of IBP is that ∫u'v dx is hopefully easier to integrate than ∫uv' dx.
eg(1) take ∫ xsin(x) dx. This is quite hard to integrate directly, so we use integration by parts. When choosing which is u(x) and v'(x), remember that you will have to integrate v' and differentiate u later. Often what happens is when you differentiate u, u'(x) turns out to be 1, which leaves you with a simple integration.
so take u(x) =x and v'(x)=sin(x)
=>u' =1, v = -cos(x) (don't worry about the "+c", it's included at the end)
so following the formula: ∫xsin(x)dx = -xcos(x) - ∫-cos(x) = xcos(x) + ∫cos(x)dx = xcos(x) + sin(x) + c
this is the final answer to that particular question, and we see that integration by parts gives us another method of integtration
1) for ∫ f(x) dx, choose suitable functions u(x) and v'(x) such that f(x) = u(x) * v'(x).
2) determine u'(x) and v(x) by differentiating and integrating respectively
3) use the formula ∫uv' dx = uv - ∫u'v dx to find the answer!
DERIVATION: (not usually necessary for exam but interesting to see!)
differentiation by parts works like this, for, u & v as functions of x,
where u' = du/dx etc.
d/dx(uv) = (uv)' = uv' + u'v (proof omitted)
if we integrate both sides wrt x
=> uv = ∫uv' dx + ∫u'v dx
=> ∫uv' dx = uv - ∫u'v dx