# Mechanics question - Maximising horizontal distance

Any advice woud be much appreciated.

The question states: 'A stone is thrown at an angle theta to the horizontol from a cliff 100m above the sea at 30m/s. At what angle should it be thrown to maximise the horizontol distance it travels?'

Now i understand what you have to do, but the algebra gets ridiculous...

x = Vtcos&#952; i + (Vtsin&#952; - 1/2gt²) j (where x = displacement, V = start speed, &#952; = angle thrown, g = acceleration due to gravity = 9.8 , t = time after throw , i and j are vectors)

Therefore, using just the 'j' component of displacement,

-100 = 30tsin&#952; - 1/2gt² (minus 100 because the stone starts 100m above it)

and solving for t using quadratic formula, gives

t = 10/g (3sin&#952; + (9sin²&#952; + 2g)^½)

Now we use the displacement vector for just i

w = 30cos&#952; t (from first equation), where w = horizontol distance.

=> w = 30cos&#952; (10/g (3sin&#952; + (9sin²&#952; + 2g)^½)) (subbing for t)

now we have to differentiate this to find the maximum, and set the differentiated bit to 0, and solve for &#952;. YEEESSSSSHHHH..i've done the differentiation and i've tried so many different ways of solving for &#952;. Can anyone help, or can anyone suggest an alternative?

In case anyone out there can be bothered to try this, after the differentiation you get:

3cos2&#952; - sin&#952;(9sin²&#952; + 2g)^½ + 9cos²&#952;sin&#952;(9sin²&#952; + 2g)^-½ = 0

If anyone can solve that for theta, i'll buy them a pint...

Cheers

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i think there is a standard result for the maximum horizontal distance and the answer is 45 degrees. i'm not sure how to prove it but i'll try anyway.

first we find what the time of flight is. let this be T.
vertical components: speed(u)=Usin(alpha), acceleration(a)=-g, displacement(S)=0, time(t)=T
using the eq'n S=ut + 1/2at^2,
Usin(alpha)T - 1/2gT^2=0
factorising, T=0 or T=2Usin(alpha)/g

now for the horizontal distance:
horizontal components: u=Ucos(alpha), S=R, t=T
using the eq'n S=ut,
R= Ucos(alpha)T
= 2U^2sin(alpha)cos(alpha)/g
simplying using the double angle formula,
R= U^2sin2(alpha)/g

for the horizontal distance to be maximum, sin2(alpha) must be maximum.
sin2(alpha)=1
therefore 2(alpha)=90
and alpha=45

and this is a standard result.
this looks very messy especially since i don't know how to use symbols to indicate angles..sorry!
basically how i did it was:
a) find time of entire journey
b) find horizontal range and substitute expression in a) for time.
c) find angle for range to be maximum.

hope this helps!
That is only true when the projectile stops at the same level from where it is thrown, here it is stopping 100m below where it starts from.
I've been trying this one for ages, and have to admit I am stumped. I can solve it using programs like mathematica. But doing it analytically is doing my head in. I get a quartic in theta squared. Which is just nasty. I guess I am missing some kind of trigonometric substitution along the line somewhere which greatly simplifies things .
Looks nasty!

Looks like a lot of manipulation is needed!

Don't look at me!

FERMAT!!!
[u="time" ="" initial="initial" speed,="speed," &#952;="&amp;#952;" required="required" angle,="angle," x="x" horizontal="horizontal" distance,="distance," y="y" vertical="vertical" t="t"]

In order for the particle to reach maximum x, it will need to be in the air for the maximum amount of time.

y = utsin&#952; - 0.5gt²

Now, we need to obtain the time the particle strikes the ground. That is, the first time y=0:

0 = utsin&#952; - 0.5gt²
t(usin&#952; - 0.5gt) = 0
t=0 or t=2usin&#952;/g (maximum t)

Then:
x = utcos&#952; = (2u²sin&#952;/g) * cos&#952; = u²2sin&#952;cos&#952;/g, since x is maximum when t is maximum.
But, 2sin&#952;cos&#952; = sin2&#952;
x = u²sin2&#952;/g

x = 900sin2&#952;/g

Since -1<sin2&#952;<1, x is max when sin2&#952;=1:
sin2&#952; = g/900
2&#952; = 0.62
&#952; = 0.31

I guess I messed up somewhere...
Yes, you forgot that the particle starts 100m above the "ground". And it simply isn't true that the particle has to be in the air for the longest time to to travel the furthest. If you directed it vertically upwards then it would be in the air for the longest time. But not travel horizontally at all.
I think I found my mistake. I overlooked the fact that the stone was 100m above sea level. So, the first time the ball strikes the ground isn't when y=0, it's when y=-100. That is:

100 = utsin&#952; - 0.5gt²

Solve that to get two values for t. Use both values in v=utcos&#952;, and continue normally.

I hope I got it right this time.
AntiMagicMan
And it simply isn't true that the particle has to be in the air for the longest time to to travel the furthest. If you directed it vertically upwards then it would be in the air for the longest time. But not travel horizontally at all.

True, but in this case I think what I did applies?
If you find the value of t in terms of theta, put that into the equation for x then you still have the problem of maximising x.

Finding x in terms of theta is easy, it is then differentiating that expression and maximising x which is the hard part.
AntiMagicMan
If you find the value of t in terms of theta, put that into the equation for x then you still have the problem of maximising x.

Finding x in terms of theta is easy, it is then differentiating that expression and maximising x which is the hard part.

Yes, it does get quite complicated.
GOTTIT!!

sin²ø = V²/[2(V² - gs)]

Where,
ø = angle of projection,
V = speed of projection,
s = distance below hill.

Just geting my claim in.
Writing it up now
Fermat
GOTTIT!!

sin²ø = V²/[2(V² - gs)]

Where,
ø = angle of projection,
V = speed of projection,
s = distance below hill.

Just geting my claim in.
Writing it up now

That doesn't make any sense. V^2 is 900, g*s is 981, so you are saying that Sin(theta) is complex.

By brute force I think that the answer is theta = 0.511076 and x = 163.6m. But that is using maple. Maybe that was a mistake by Fermat, I would be very interested in a proper proof though.
As some have noted, the working out/algebra was like to be pretty horrible, and it would be difficult to follow the working if it wasn't formatted properly.
However, it's been ages since I did any latex, so I just wrote it out and scanned it in. Hope you can read it/follow it.

I started at the eqns for the time of flight, and ended with an expression for maximising the (horizontal) distance travelled, depending on the angle of trajectory.

Oh, yes. In the formula,

sin²ø = V²/[2(V² - gs)]

upwards is measured as +ve, so s should be entered as s = -100, for example.

Edit: latex'd the proof.
Nice. Very impressive. And that gives the same theta I obtained using maple.
AntiMagicMan
Nice. Very impressive. And that gives the same theta I obtained using maple.

Ah, great. Now I feel my work is justified.
I'm afraid my mathematica has run its course though
Now I'm using sci-lab
Oooh nice, Open Source.

I gave up trying to solve for theta once I had got to an equation that had the sixth power of Sine in it :s. I haven't read your solution thoroughly but given that it gives the same answer as Maple, it is almost certainly right. I will when I get a bit of time. I guess I just need to brush up on Trig formulas.
More rep for Fermat I feels
shiny
More rep for Fermat I feels

Well i said i'd buy you a pint, Fermat....thanks for doing that.