The Student Room Group

STEP I : Mechanics question

Right, I have not done many of these questions before so I have every reason to be curfuddled.

It is Question 9 from STEP I 2004;

A particle is projected over level ground with speed u at an angle θ above the horizontal. Derive an expression for the greatest height of the particle in terms of u, θ and g

This part is fair enough

A particle is projected from the floor of a horizontal tunnel of height 9/10(d). Point P is ½d metres vertically and d metres horizontally along the tunnel from the point of projection. The particle passes through point P and lands inside the tunnel without hitting the roof. Show that;

arctan0.6 < &#952; < arctan3


With this I used the fact the angle will be at a maximum when the maximum height of the ball is basically the roof (and it worked with arctan3)

However, when I used the pretence that the maximum height at the minimum angle would be ½d, it did not work out using arctan0.6, instead working out at arctan1, which is a bit silly I presume.

I have these equations;

Max Height = u²sin²&#952;/2g

½d = dtan&#952; - ½g( /u²cos²&#952; )

Are these correct for a start ? lol

Any help much appreciated :biggrin:
Reply 1
You should use the fact that the maximum height is always less than (9/10)d, and from that inequality you should get a quadratic in tan&#952;.
Reply 2
yeah, I have a quadratic

½=tan&#952; - 1/4k(tan²&#952; ) where k is the maximum height

therefore: tan²&#952; - 4ktan&#952; + 2k = 0

if k =0.9

=> .....you've got it spot on, I made myself two quadratics for whatever reason, I will now jump in a hole and die :p: