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

Vertical height gained: \(h=2.5\sin 30^\circ=1.25\text{ m}\).

Use conservation of mechanical energy.

\(\tfrac12 mv^2 = mg\cdot 1.25\).

The mass cancels, so \(\tfrac12 v^2=12.5\).

Hence \(v^2=25\).

Therefore \(v=5\text{ m s}^{-1}\).

For the maximum possible speed, the ball bearing just comes to rest at the barrier.

Vertical height gained: \(h=1.2\sin 10^\circ\).

Use conservation of mechanical energy:

\(\tfrac12 mv^2 = mg\cdot 1.2\sin 10^\circ\).

The mass cancels, so \(v^2=24\sin 10^\circ\).

Therefore \(v\approx 2.04\text{ m s}^{-1}\).

Use conservation of mechanical energy.

\(\tfrac12 m\cdot 15^2-\tfrac12 mv^2 = mg\cdot 1\).

The mass cancels, so \(225-v^2=20\).

Hence \(v^2=205\).

Therefore \(v\approx 14.3\text{ m s}^{-1}\).

Let the launch angle be \(\theta\).

At the top of the flight, the vertical component of velocity is zero, so the speed is \(15\cos\theta\).

Use conservation of mechanical energy:

\(\tfrac12 m\cdot 15^2-\tfrac12 m(15\cos\theta)^2 = m\cdot 10\cdot 1.45\).

The mass cancels, so \(225-225\cos^2\theta = 29\).

Hence \(\cos^2\theta=\dfrac{196}{225}\approx 0.8711\).

So \(\cos\theta\approx 0.9333\).

Therefore \(\theta\approx 21.0^\circ\).

Use conservation of mechanical energy.

\(\tfrac12 mv^2 = mgh\).

The mass cancels, so \(\tfrac12 v^2 = 10h\).

Hence \(h=\dfrac{v^2}{20}\).

Use conservation of mechanical energy.

\(\tfrac12 mv^2-\tfrac12 mu^2 = mg\cdot 10\).

The mass cancels, so \(v^2-u^2=200\).

Hence \(v^2=u^2+200\).

Therefore \(v=\sqrt{u^2+200}\text{ m s}^{-1}\).