Answer:

(a) \(y=x\tan\theta-\dfrac{gx^2}{2u^2}(1+\tan^2\theta)\).

(b) There is no real value of \(\theta\) for which the particle passes through \((18,30)\).

(a) From the List of formulae, the trajectory equation is

\(y=x\tan\theta-\frac{gx^2}{2u^2}\sec^2\theta.\)

Use the identity

\(\sec^2\theta=1+\tan^2\theta.\)

Substituting this into the trajectory equation gives

\(y=x\tan\theta-\frac{gx^2}{2u^2}(1+\tan^2\theta).\)

(b) First use the fact that when \(\theta=\tan^{-1}2\), the projectile passes through \((10,16)\). Since \(\tan\theta=2\),

\(1+\tan^2\theta=1+4=5.\)

Substitute \(x=10\), \(y=16\), and \(g=10\) into the formula from part (a):

\(16=10(2)-\frac{10(10)^2}{2u^2}(5).\)

So

\(16=20-\frac{2500}{u^2}.\)

Hence

\(\frac{2500}{u^2}=4,\)

and therefore

\(u^2=625.\)

Now suppose the particle could pass through \((18,30)\). Let

\(T=\tan\theta.\)

Using \(u^2=625\), the trajectory equation gives

\(30=18T-\frac{10(18)^2}{2(625)}(1+T^2).\)

Since

\(\frac{10(18)^2}{2(625)}=\frac{3240}{1250}=\frac{324}{125},\)

we get

\(30=18T-\frac{324}{125}(1+T^2).\)

Multiply by \(125\):

\(3750=2250T-324(1+T^2).\)

Rearrange:

\(324T^2-2250T+4074=0.\)

For this quadratic to have a real solution for \(T\), its discriminant must be non-negative. The discriminant is

\(\Delta=(-2250)^2-4(324)(4074).\)

Calculating,

\(\Delta=5062500-5279904=-217404.\)

This is negative. Therefore the quadratic has no real solution for \(T=\tan\theta\), and hence there is no value of \(\theta\) for which \(P\) can pass through \((18,30)\).