projectile question (1 Viewer)

[sasquatch]

I'm having trouble getting the answer i the back of the book.. If anyone has done this question, could they please check for me>

It's Cambrdige Exercise 7.3 Q11b

The answer says 10(root2 - 1) ms^-1, but i get 5(2root2 - 3) ms^-1...

Yeah ive done the question twice..keep getting that answer..

Thanks for any help..

 

[gamecw]

yo type the question?

 

[Riviet]

Cambridge excercise 7.3

11 A projectile is fired from a point O on level ground with speed 13 m/s at an angle of projection alpha, where tan(alpha) = 12/5. The projectile just clears the top of a wall in its path and then reaches a maximum height of twice the height of the wall. At the instant of projection, a target is fired horizontally from the top of the wall and continues to move horizontally with constant speed u in the plane of the path of the projectile away from O. Find

(a) the distance of the base of the wall from O
(b) the value of u, given that the projectile hits the target

 

[sasquatch]

Oh sorry.. nuh i didnt really intend for someone to actualyl do it for me..i appreciate that.. just i guessed that people would have had to already have done it..so yeah just wondered if someone could check their answer..

Thanks for typing it up riviet..

 

[haque]

LOL seeing as i'm sick and have nothing to do here's the answer
the equation for the displacement of the horizontally projected particle is
d=ut +6-3sqrt2 and from the previous part we know that for y=h, x=6+3sqrt2 for the downward flight(its a parabola so its symmetrical) and so when x=6+3sqrt2 t=6+3sqrt2/5 using x=vtcos@. sub this into the "d" equatioon above and simplify and u get the answer at the back.