Projectiles (1 Viewer)

[AE86]

I was wondering if we are allow to make an assumption that the position of max height is also the midpoint of the horizontal range providing that the particle is projected at the height 0 and hitting the ground at 0.

Thanks

 

[Xayma]

Given that it forms a parabola, if one of the x-intercepts of the parabola is x=0, then yes the max height will be half horizontal range, although I dont think you could make that assumption without justification (ie stating that it forms a parabola and since parabolas are symmetrical it is halfway), may I ask why you would figure out the x-component of max height, just take the time value and double it.

 

[AE86]

Apparently, this is the question i was doing.

A projectile reaches a maximum height of 100m above the point of projection at a horizontal distance of 250m from this point. Show that it was fired at an angel of elevation of approximately 38 degrees and 40' and calculate the maximum range for the same velocity.

So, i was wondering if i can just assume that the range of would be 500m.

 

[Li0n]

Apparently, this is the question i was doing.

A projectile reaches a maximum height of 100m above the point of projection at a horizontal distance of 250m from this point. Show that it was fired at an angel of elevation of approximately 38 degrees and 40' and calculate the maximum range for the same velocity.

So, i was wondering if i can just assume that the range of would be 500m.

I wouldn't.

 

[AE86]

So how would i go about with this question?

Would this be a typical HSC question?

This question is from some blue 3unit text book by S.B. Jones. It's old and the examples it had didn't even involve any calculus but brute formulas.

I just need ideas, i dont really need full working.

THanks

 

[Xayma]

Derive the cartesion and show that it works for y=100m and x=250m at θ=38° and V=40.

You could also show that with those values they have the same time. From that do the normal max range ie you could do symmetry of a parabola or y=0 find t and then sub it in x displacement.

 

[Steven12]

do the following


Height=(VsinQ)Square / 2g (u should know this equation by now)
which is 100= (VsinQ)square / 19.6
rearrange 1960=(vsinQ)square
rearrange Root(1960)=VsinQ
so VsinQ=44.5

then we work out the half time

t=vsinQ(44.5)/9.8
t=4.54(s)

once we have the half time, sub it into the horizontal componet equation

x=vcosQt (since we know the half time, and the half distance.)

250=VcosQ(4.54)
vcosQ=55.1(approx) always keep exact value in your calculator

so now VsinQ=44.5
VcosQ=55.1
tanQ=0.81
Q=38.9'

there you go

with max range. all u need to do is to double the time. sub it into the horizontal componet equation and its allllllllll smooth

one thing im not sure about is if u need to derive the equation if the question doesnt ask for it. does anybody know?

 

[mojako]

I wouldn't.

I would.
especially if the question doesn't have other parts that lead you to finding the range the hard way, and it doesn't have ridiculously high marks (eg 5). But I would say:
the path of a projectile [which is the object thrown] is always parabolic in shape. the projectile in this question starts and stops moving at the same vertical altitude.
maximum height is reached at x = 250, so (250, max height) corresponds to the vertex of the parabola.
x = 250 is the middle of the horizontal range.
hence the range is 500 m.

 

[Sirius Black]

Does it hav to start at the origin?

 

[mojako]

Does it hav to start at the origin?

if it doesnt start at the origin, you can consider the inverted U path only, i.e. don't look atthe extra part.. I hope I'm saying this clear enough.. I could actually draw it...

Taking the original question by AE86:
"I was wondering if we are allow to make an assumption that the position of max height is also the midpoint of the horizontal range providing that the particle is projected at the height 0 and hitting the ground at 0"

The position of max height will not be on the midpt of the horizontal range.
But.. it will be the midpoint of.. well... I'm sure you know, otherwise I'll send you a picture.