Hi!
I'm having a little trouble with this question from the Cambridge 3U textbook (Q10 ex. 3H). Can anyone help with part c of this question?
A particle is projected from the origin with velocity V m/s at an angle of a to the horizontal.
a) Assuming that the coordinates of the particle at time t are (Vtcosa, Vtsina - 1/2gt2, prove that the horizontal range R of the particle is [V2sin2a]/g
b) Hence prove that the path of the particle has equation y = x (1 - x/R) tana
c) Suppose that a=45o and that the particle passes through two points 6m apart and 4m above the point of projection. Let x1 and x2 be the x-coordinates of the two points.
Thank you so much!
I'm having a little trouble with this question from the Cambridge 3U textbook (Q10 ex. 3H). Can anyone help with part c of this question?
A particle is projected from the origin with velocity V m/s at an angle of a to the horizontal.
a) Assuming that the coordinates of the particle at time t are (Vtcosa, Vtsina - 1/2gt2, prove that the horizontal range R of the particle is [V2sin2a]/g
b) Hence prove that the path of the particle has equation y = x (1 - x/R) tana
c) Suppose that a=45o and that the particle passes through two points 6m apart and 4m above the point of projection. Let x1 and x2 be the x-coordinates of the two points.
i) show that x1 and x2 are the roots of the equation x2 - Rx + 4R
ii) Use the identity (x1 - x2)2 = (x1 - x2)2 - 4x1x2 to find R
Thank you so much!