CM_Tutor leaving (2 Viewers)

[CrashOveride]

Dont you mean tan@ - t / (6cos@)?

EDIT: Hmm simplyifing early on made it a lot easier...thnx BlackJack

 

[CM_Tutor]

CrashOveride, I had a quick look at the question with the circle, but it doesn't make sense to me. Taking the circle as x² + y² = U⁴ / g², then for the missile to hit (0, U² / g), it must be fired vertically. However, I get that the max height when fired vertically is U² / 2g, and so it only gets half way.

Sorry, but I don't have time to sort this out. Will look when I get back if it's still in issue.

ALL: See you when I get back, and please leave me a note here about anything you want me to have a look at - just identifying thread and post number would be great. Thanks.

 

[BlackJack]

(I believe) The problem with the question here, CM, was the double definition of the x,y-axis... you have to show that the max range is U^2/g first, but the coordinates you use for the missile's trajectory isn't what they were asking for in the second part. To follow the question, P(x,y) must lie flat on the ground on the x,y-axes and the missile's path must be on a different set of axes so that it flies through the air, and hits the maximum at the edge of the circle when height = 0

The explanation file I gave Crash is attached here.

 

[CrashOveride]

A stone is projected with an initial velocity v1 in a vertical plane at an angle ¥ to the horizontal and hits ground at A. another stone launched with initial veloc. v₂ at same angle but v₂ > v₁. Let B(x2, y2) and C(x1,y1) be two points on the respective flight paths at the same time t. Show that gradient of segment BC is independant of time t. <B>Done</B>

When 2nd stone just clears a wall of hieght h, the first stone hits the ground at A. If the wall stands at point D on the level ground, prove that AD = hcot¥. <B>Done</B>

Further show that tan(-Ø) = tan¥ - [gt / (v2 cos¥)], where Ø is the angle made by the downward flight of the faster stone with the horizontal and T is the time of flights for the slower stone. Hence show v₂(tan¥ + tanØ) = 2v₁tan¥

EDIT: Ok i can show that tan(-Ø) = tan¥ - (gt/(V₂cos¥)) which is kind of close, just 't' is not the time of flight of the slower stone, its just a time 't'. Hrmm

Actually i get the exact answer, except my time is just a time 't' where as its supposed to be the time of flight 'T" of the smaller stone?? And to do the hence thing at the end i have to substitute the time of flight T in for the thing to work out