Another projectile motion (1 Viewer)

[HeroicPandas]

A ball is projected from a point on the ground, distance a from the foot of a vertical wall of height b. The ball is projected at an elevation with speed V.

a) Find how high above the wall the ball passes it.

b) If the ball just cleans the wall, prove the greatest height reached is:

 

[HeroicPandas]

I just need help in part b), part a) is used for a guide to do part b)

 

[Sy123]

Ok so we find the cartesian equation of motion of the projectile and sub in x=a y=b:



Now we are asked to find the greatest height given the restriction above.

The greatest height in general for ANY V and any theta will be (after finding subbing into y etc)
We find:



Ok look at the expression we are trying to prove, if you notice, V^2 is absent from that expression, its a hint to use our given restriction in terms of V^2 and sub it into y max. When I say restriction, our parabola of motion is restricted in V and in theta such that it passes through (a,b). (Note just because I say restriction doesnt mean that they cant vary).

So lets transform our first expression with V^2 as the subject:



So we now have our V^2, the special thing about this V^2 is that it is such that it will grant us motion through (a,b).
So lets sub it into y-max, its difficult to latex the algebra, so what you do is sub in that V^2 into y_max you should get various cancellations and simplifications to get your answer

 

[HeroicPandas]

Thank you very much Sy