Q. motion (1 Viewer)

[marsenal]

These are from Patels book,

1.The nozzle of a water hose is at a point O on the horizontal sround. the water comes out of the nozzle with speed U m/s. Neglecting the air resistance, prove that the water can reach the wall at a distance d from O, if U²>gd, where g is the acceleration due to gravity.
If U²=4gd, also prove that the maximum height that can be reached by the jet on this wall is given by 15d/8.

For this one I get up to gd=U²sin2@, but am not sure what to state then to prove the first part.

2.A missile is fired from O with initial velocity U at an angle @ with the horizontal. Prove that it describes a parabola of focal length U²cos²@/(2g).
Also prove that any point P(x,y) within and on the cirle x²+y²=v⁴/g² is in danger of being hit by the missile (g/m/s/s is the acceleration due to gravity)

I can get the first part of this question but don't know where to go from there.

 

[wogboy]

1.

let y be the vertical displacement above the ground, x be be the horizontal displacement to the left of the starting point, at time t:

start off with the vertical displacement y:
d^2(y)/dt^2 = - g
dy/dt = Usin@ - gt (since at t=0, dy/dt = Usin@)
y = Utsin@ - g(t^2)/2

from this information, lets find T, the time of flight:
0 = UTsin@ - g(T^2)/2
since T =/= 0, we find the non zero value of T in this quadratic:
T = (2Usin@)/g

now for the horizontal displacement:
dx/dt = Ucos@
x = Utcos@ (when t=0, x=0)

if we sub in t=T into the equation for horizontal displacement, in order to find the horizontal range R:
R = UTcos@
R = U(2Usin@cos@)/g
R = (U^2)(sin2@)/g

now we need to know when the range R is maximised, so we know at what angle @ to shoot the water out at, for most efficiency to reach the wall. The maximum value sin2@ can give is 1 for any real angle @, and this occurs at @=45 deg=pi/4. So the maximum range R is:

R = (U^2)/g (only when the water is shot out at @=45 deg)

You should know from instinct that when projecting a particle, maximum range is achieved at @=45 deg. Now we want the water to reach the wall, where x=d. So the range should be at least the distance of the wall:

therefore R>=d.
(U^2)/g >= d
U^2 >= gd
Note that this is only when @=45 deg, at optimum efficiency. Whenever U^2 is any lower, there's no possible way the water can reach the wall.

To find out when the water hits the wall:

x = d,
therefore d = Ut*cos@
t = d/Ucos@

at this value of t, we need to find y:
y = Utsin@ - g(t^2)/2
y = dtan@ - { g(d^2)sec^2(@) / 2U }

however, U = 4gd
y = dtan@ - d{sec^2(@)}/8
y = d[tan@ - (1/8)sec^2(@)]
y = d[tan@ - (1/8)tan^2(@) - (1/8)]

to find the maxiumum possible value of y, we differentiate WRT @ and equate to zero, to find the value of @ that optimises y:
dy/d@ = 0
sec^2@ - (1/4)tan^2(@)*sec^2(@) = 0
sec^2(@) = 0 or tan@ = 4
since sec^2(@) has no solutions,
tan@ = 4, gives the optimal value of @.
subbing this back in to find y:

y = (15/8)*d
so the maximum height the water can ever reach when U=4gd is (15/8)*d, and this is only when @ = arctan(4) (optimal value of @)

 

[wogboy]

2.

Also prove that any point P(x,y) within and on the cirle x^2+y^2=v^4/g^2 is in danger of being hit by the missile

Where did the variable v come from?

 

[ND]

Originally posted by wogboy

Where did the variable v come from?

I'd guess that v is supposed to be U. If v were a variable, then x^2+y^2 = v^4/g^2 wouldn't be a circle.

 

[marsenal]

Thanks for the solution to the first one. I appreciate you not missing any steps.
In regard to the second one and the "v", that wasn't a typo, and I'm assuming that it's referring to velocity at the particular point (x,y).