projectile motion (1 Viewer)

[jkwii]

i got a Q from patel. it looks easy but maybe i'm having a bad day.

the nozzle of a water hose is at a point O on the horizontal ground. the water comes out of the nozzle with speed U m/s. neglecting air-resistance, prove that the water can reach a wal at a distance d from O, if U^2 . gd where g is the acceleration of gravity.

If U^2 = 4gd, also prove that the max height that ca be reached on this wall is given by 15d/8.

 

[lolokay]

x = angle of projection, t = time

d = Ucosx t
v = Usinx - gt, where v = -Usinx (same speed in opposite direction at end of flight_
t = 2Usinx/g
d = Ucosx 2Usinx/g
= U^2 sin2x/g
d' = U^2/g * cos2x = 0 , at maximum distance
cos2x = 0, x = 45 at max d
so max d = U^2/g
and if U^2 = gd
d = gd/g = d
so the water can reach the (bottom of the) wall


if U^2 = 4gd, h = height

d = Ucosx t
t = d/Ucosx
h = Usinx t - 1/2 gt^2
= dUsinx/Ucosx - 1/2 gd^2/U^2cos^2x
= dtanx - dU^2/8U^2cos^2x
= d(tanx - 1/8(cos^2 x)
h' = d(1/cos^2x - 2sinx/8cos^3x)
= d(4cosx/4cos^3x - sinx/4cos^3x) = 0 (at maximum)
4cosx - sinx = 0
tanx = 4
so cos^2x = 1/17

put it back into height formula: h = d(tanx - 1/8(cos^2 x)
h = d(4 - 1/8*(1/1/17)
h = d15/8

 

[jkwii]

that was really good!! thanks a heap.