Answer:

(a) \(T=2\). (b) The direction is \(\tan^{-1}\!\left(\frac{8}{9}\right)\), about \(41.6^\circ\) below the horizontal.

Since

\(\tan\theta=\frac43,\)

we can use the \(3\)-\(4\)-\(5\) triangle:

\(\sin\theta=\frac45,\qquad \cos\theta=\frac35.\)

Let the collision occur after \(T\) seconds.

The model aircraft starts from a point vertically above \(P\), which is \(8\) m horizontally from \(O\). It then travels horizontally at \(5\text{ m s}^{-1}\). Therefore, at time \(T\), its horizontal distance from \(O\) is

\(8+5T.\)

The horizontal displacement of the ball after time \(T\) is

\(u\cos\theta\,T.\)

At collision, these horizontal distances are equal:

\(u\cos\theta\,T=8+5T.\)

Using \(\cos\theta=\frac35\),

\(\frac35uT=8+5T.\)

The aircraft remains at height \(4\) m above the ground, so the vertical displacement of the ball at collision is \(4\) m. Hence

\(u\sin\theta\,T-\frac12gT^2=4.\)

Using \(g=10\) and \(\sin\theta=\frac45\),

\(\frac45uT-5T^2=4.\)

From the horizontal equation,

\(u=\frac{5(8+5T)}{3T}.\)

From the vertical equation,

\(\frac45uT=4+5T^2,\)

so

\(u=\frac{5(4+5T^2)}{4T}.\)

Equate these two expressions for \(u\):

\(\frac{5(8+5T)}{3T}=\frac{5(4+5T^2)}{4T}.\)

Cancel the common factor \(\frac{5}{T}\):

\(\frac{8+5T}{3}=\frac{4+5T^2}{4}.\)

Cross-multiply:

\(4(8+5T)=3(4+5T^2).\)

So

\(32+20T=12+15T^2.\)

Rearrange:

\(15T^2-20T-20=0.\)

Divide by \(5\):

\(3T^2-4T-4=0.\)

Factorise:

\((3T+2)(T-2)=0.\)

Since time must be positive,

\(T=2.\)

Now find \(u\) using the horizontal equation:

\(\frac35u(2)=8+5(2).\)

So

\(\frac65u=18,\)

and therefore

\(u=15.\)

The horizontal velocity of the ball is constant:

\(v_x=u\cos\theta=15\cdot\frac35=9.\)

The vertical velocity immediately before collision is

\(v_y=u\sin\theta-gT.\)

Substitute \(u=15\), \(\sin\theta=\frac45\), \(g=10\), and \(T=2\):

\(v_y=15\cdot\frac45-10(2)=12-20=-8.\)

The negative sign means the ball is moving downwards.

If the angle below the horizontal is \(\phi\), then

\(\tan\phi=\frac{|v_y|}{v_x}=\frac89.\)

Therefore the direction of motion is

\(\tan^{-1}\left(\frac89\right)\)

below the horizontal.

Numerically, this is approximately

\(41.6^\circ\)

below the horizontal.