Easy M2 question

The unit vectors I and J are horiztonal and vertically upwards respectively. A particle is projected with velocity (8i + 10j)ms-1 from a point O at the top of a cliff and moves freely under gravity. Six seconds after projection, the particle strikes the at the point S. Calculate:

(a) the horizontal distance between O and S
(b) the vertical distance between O and S
(c) At time T seconds after projection, the particle is moving with velocity (8i - 14.5j)ms-1. Find the value of T and the position vector, relative to O, of the particle at this instant

Cant do c, any help please..

Bump

The vertical component of velocity has changed by 24.5 m/s. Acceleration is 9.8 m/s^2. So T = 24.5/9.8 = 2.5.

Position vector at time T in metres
= (8T, (1/2)(10 + 10 - 9.8T)T)
= (20, -5.625).

Reply 3

Barny

Jonny W

The vertical component of velocity has changed by 24.5 m/s. Acceleration is 9.8 m/s^2. So T = 24.5/9.8 = 2.5.

Position vector at time T in metres
= (8T, (1/2)(10 + 10 - 9.8T)T)
= (20, -5.625).

Ahh, stupid me, nice one.

Reply 4

diablo2004

a = -gj

v = ∫a dt = -gt[j] + C

t=0, v = 8i + 10j => C = 8i + 10j => v = 8i + (10-gt)j

r = ∫v dt = 8ti + (10t-gt/2)j + C

t=0, r = 0 => C = 0 => r = 8ti + (10t-gt/2)j

a) t=6 => r = 48i + (80-3g)j

|r_horiz| = 48m

b) |r_vert| = (80-3g) m

c) 8i + (10-gT)j = (8i - 14.5j) => T = 24.5/g

r = 8Ti + (10T-gT/2)j = 196/g i + (245/g - 12.25)j

edit: crap. i spent 15 minutes making all the vectors bold :rolleyes: