Melody, your 2nd answer is correct!

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If you move your eyes straight to the bottom you will see that I am not comfortable with what I have done!

So you should probably  wait for someone else.

For part A

You would need to know the angle incline of the slope.

Because you need to have the vertical component of the initial velocity 

and that would be 96sin(angle)        

I suppose that you would use 2.5 degrees

$$V_y=96sin(2.5^0) \approx 4.187461187 km/hour$$  

$${\frac{{\mathtt{4.187\: \!461\: \!187}}{\mathtt{\,\times\,}}{\mathtt{1\,000}}}{\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{60}}\right)}} = {\mathtt{1.163\: \!183\: \!663\: \!055\: \!555\: \!6}}$$

approx vertical velocity is 1.1632 m/s   Initially   u=1.1632

a = -9.8m/s^2

Find height (s) when velocity = 0      v=0

Number 4 has u, a, v,    and s so use that one

$$\\v^2=u^2+2as\\\\ 0=1.16318^2+2*-9.8*s$$

$${\mathtt{0}} = {{\mathtt{1.163\: \!18}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}\left({\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{s}}\right) \Rightarrow {\mathtt{s}} = {\frac{{\mathtt{114\,603\,045}}}{{\mathtt{1\,660\,192\,226}}}} \Rightarrow {\mathtt{s}} = {\mathtt{0.069\: \!029\: \!985\: \!326\: \!530\: \!5}}$$

This is the vertical height that it will coast in metres.

Draw the triangle and you can work out the distance it will coast up the hill

$${\frac{{\mathtt{0.069}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{2.5}}^\circ\right)}}} = {\mathtt{1.581\: \!865\: \!408\: \!209\: \!866\: \!1}}$$

1.58 metres - That does not look right  

HELP - What did I do wrong ??

»You would need to know the angle incline of the slope

Not when the question asks merely how high the hill

All the car's KE is converted to gravitational potential energy

½mv² = mgh

∴ h = ½v² ÷ g

DONE

part (b)

It doesn't gain as much gravitational PE as it would if losses were absent

so friction losses = mgh_{ideal -}  mgh_actual

I'm not sure what I did to get Courier text and all those blank lines.

☹

Thanks you anon 

ok so 

96km/hour

$${\frac{{\mathtt{96}}{\mathtt{\,\times\,}}{\mathtt{1\,000}}}{\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{60}}\right)}} = {\frac{{\mathtt{80}}}{{\mathtt{3}}}} = {\mathtt{26.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

$$\begin{array}{rll} KEnergy&=&gravitational\; Potential \;energy\\ 0.5mv^2&=&mgh\\ h&=&0.5v^2/g\\ h&=&0.5*(26.\dot6)^2/9.8 \;m\\ h&=&0.5*711.111/9.8\;m\\ h&\approx&36.28m \end{array}$$

Is this correct?  It does sound better.  

If this one is correct then what was wrong with my logic the first time?

Melody, your initial logic (the use of v² = u² + 2as) would work if you imagine the car going straight up:

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Thanks Alan,

I thought I was just taking the upward component.  I cannot do that I am supposing. :)

Is my second answer correct?

Thanks Alan