IAL Edexcel Jan 2019 Mech2 Q8b(ii) & 8c

Hello, I will appreciate help on these two question.

This is the link to the paper. https://qualifications.pearson.com/content/dam/pdf/International%20Advanced%20Level/Mathematics/2013/Exam%20materials/WME02_01_que_20190123.pdf

8b(ii) I haven't been able to understand the question nor have I understood the markscheme. I assumed that since the object was falling freely under gravity then it's direction of motion would be just vertically downwards.

8c) I've used the formula V^2=u^2+2as taking v as 0 and u as 15 but I'm wrong

All help is appreciated.

Not worked it through but the particle has speed 10m/s as it leaves B in the direction of the ramp so that provides both horizontal and vertical initial velocities. So its not just vertical motion. But using the KE/PE conservation of energy with the "initial" speed of 10 should give w that that initial KE.

The greatest height would be by resolving the "initial" speed 10m/s and analysing the vertical component

Thank you I still don't get the part b(ii) but I have worked out s and got 0.05968... however the markscheme gives the final s value as 0.05968+6sin20. Why should the 6sin20 be added

For part c I think this sketch has made me understand it quite well now.I get why we need the 6sin20. Is my value of 0.0569 accurately placed?
For part b(ii) this is my sketch. I can't understand why we are using 10sin20 as the horizontal distance and not 11.8costheta. I also understand that I wouldn't be able to get a value using my sketch.

For c) that looks ok. A sketch should make it clear that you sum the 0.5 and the B height to get the height above ground.

For bii) the direction of motion as it hits the ground. You know the horizozntal velocity is unchanged from the time it leaves B to when it hits the ground. So 10cos(20) must be equal to w*cos(theta) which gives theta.