Use conservation of mechanical energy.

Vertical height lost: \(h=3.5\sin 20^\circ\).

So \(\tfrac12\cdot 3\cdot 8^2-\tfrac12\cdot 3\cdot u^2=3\cdot 10\cdot 3.5\sin 20^\circ\).

This gives \(96-1.5u^2=35.91\).

Hence \(1.5u^2=60.09\), so \(u^2=40.06\).

Therefore \(u\approx 6.33\text{ m s}^{-1}\).

Use conservation of mechanical energy.

\(\tfrac12 mv^2 = mg\cdot 1.43\).

The mass cancels, so \(\tfrac12 v^2=14.3\).

Hence \(v^2=28.6\).

Therefore \(v\approx 5.35\text{ m s}^{-1}\).

Convert mass to kilograms: \(57\text{ g}=0.057\text{ kg}\).

Increase in GPE \(=mgh=0.057\times 10\times 1\).

So the increase in gravitational potential energy is \(0.57\text{ J}\).

First convert \(180\text{ km h}^{-1}\) to SI units.

\(180\text{ km h}^{-1}=50\text{ m s}^{-1}\).

The gain in potential energy is \(0.057\times 10\times 1=0.57\text{ J}\).

Use conservation of energy:

\(\tfrac12\cdot 0.057\cdot 50^2-\tfrac12\cdot 0.057\cdot v^2=0.57\).

This simplifies to \(50^2-v^2=20\).

So \(v^2=2480\), hence \(v\approx 49.8\text{ m s}^{-1}\).

At the top of the flight the velocity is horizontal, so the horizontal speed is \(49.8\text{ m s}^{-1}\).

Convert \(20\text{ cm}\) to metres: \(20\text{ cm}=0.2\text{ m}\).

Use conservation of mechanical energy.

\(\tfrac12 mv^2 = mg\cdot 0.2\).

The mass cancels, so \(\tfrac12 v^2=2\).

Hence \(v^2=4\).

Therefore \(v=2\text{ m s}^{-1}\).