oh no, mechanics.

this is my first homeowrk question i'm asking for help on. i feel so dirty :rolleyes:

a stone is thrown vertically upwards with a speed of u metres per second. a second stne is thrown vertically upwards from the same point with the same initial speed but T seconds later. prove they collide at a distance of

\frac{4u^2 - g^2T^2}{8g}

m above the point of projection.

Reply 1

steve2005

17

chewwy

this is my first homeowrk question i'm asking for help on. i feel so dirty :rolleyes:

a stone is thrown vertically upwards with a speed of u metres per second. a second stne is thrown vertically upwards from the same point with the same initial speed but T seconds later. prove they collide at a distance of

\frac{4u^2 - g^2T^2}{8g}

m above the point of projection.

I got halfway through before I read the whole question: so I need to make some adjustments.

See what I have done so far.

chewwy

this is my first homeowrk question i'm asking for help on. i feel so dirty :rolleyes:

a stone is thrown vertically upwards with a speed of u metres per second. a second stne is thrown vertically upwards from the same point with the same initial speed but T seconds later. prove they collide at a distance of

\frac{4u^2 - g^2T^2}{8g}

m above the point of projection.

I have the distinct impression that you have given at least as much as you've got on this forum!

Feel good about it!

Aitch

[You should have seen how many questions I posted when I was strugggling to teach myself FP2... :eek:

]

Reply 3

Fermat

8

Let s1 be the distance travelled by the first stone.
Let s2 be the distance travelled by the second stone.

s1 = ut - ½gt²
s2 = u(t-T) - ½g(t-T)²

the stones collide when s1 = s2

ut - ½gt² = u(t-T) - ½g(t-T)²
ut - ½gt² = ut - uT - ½gt² + gtT - ½gT²
uT + ½gT² = gtT
t = (u+½gT)/g
==========

substituting for t into the eqn for s1,

s1 = u(u + ½gT)/g - ½g(u + ½gT)²/g²
s1 = {u² + ½guT - ½u² - ½guT - (1/8)g²T²}/g
s1 = {½u² - (1/8)g²T²}/g
s1 = (4u² - g²T²)/(8g)
================

Reply 4

Chewwy

OP

17

Fermat

Let s1 be the distance travelled by the first stone.
Let s2 be the distance travelled by the second stone.

s1 = ut - ½gt²
s2 = u(t-T) - ½g(t-T)²

the stones collide when s1 = s2

ut - ½gt² = u(t-T) - ½g(t-T)²
ut - ½gt² = ut - uT - ½gt² + gtT - ½gT²
uT + ½gT² = gtT
t = (u+½gT)/g
==========

substituting for t into the eqn for s1,

s1 = u(u + ½gT)/g - ½g(u + ½gT)²/g²
s1 = {u² + ½guT - ½u² - ½guT - (1/8)g²T²}/g
s1 = {½u² - (1/8)g²T²}/g
s1 = (4u² - g²T²)/(8g)
================

thankyou.

Reply 5

steve2005

17

Fermat

Let s1 be the distance travelled by the first stone.
Let s2 be the distance travelled by the second stone.

s1 = ut - ½gt²
s2 = u(t-T) - ½g(t-T)²

the stones collide when s1 = s2

ut - ½gt² = u(t-T) - ½g(t-T)²
ut - ½gt² = ut - uT - ½gt² + gtT - ½gT²
uT + ½gT² = gtT
t = (u+½gT)/g
==========

substituting for t into the eqn for s1,

s1 = u(u + ½gT)/g - ½g(u + ½gT)²/g²
s1 = {u² + ½guT - ½u² - ½guT - (1/8)g²T²}/g
s1 = {½u² - (1/8)g²T²}/g
s1 = (4u² - g²T²)/(8g)
================

I wish I could produce such elegant solutions.

I won't bother to post my effort. By-the-way I have replaced the 2 by T. I must not be in so much of a hurry.

Reply 6

username9816

14

chewwy

this is my first homeowrk question i'm asking for help on. i feel so dirty :rolleyes:

a stone is thrown vertically upwards with a speed of u metres per second. a second stne is thrown vertically upwards from the same point with the same initial speed but T seconds later. prove they collide at a distance of

\frac{4u^2 - g^2T^2}{8g}

m above the point of projection.

For stone 1:
x = ut - (g/2)t^2 = t(u - gt/2)

For stone 2:
x = u(t + T) - (g/2)(t^2 + 2tT + T^2)

At POC:
-> ut - (g/2)t^2 = ut + uT - gt^2/2 - gtT - gT^2/2
-> uT - gtT - gT^2/2 = 0
-> u - gt - gT/2 = 0
-> gt = u - gT/2
-> t = (u/g) - T/2

Hence this expression for time t satisfies both equations of x for Stone 1 and Stone 2. Sub. t = (u/g) - T/2 into the equation of x for Stone 1 gives:

x = t(u - gt/2) = [(u/g) - T/2]{u - g[(u/g) - T/2]/2} = [(2u - gT)/2g]{u - g[u/g - T/2]/2} = [(2u - gT)/2g]{u - (2u - gT)/4]} = [(2u - gT)/2g][(2u + gT)/4] = (2u - gT)(2u + gT)/(8g) = (4u^2 + g^2T^2)/(8g)

Where x is the distance above the point of projection.

Reply 7

steve2005

17

Nima

For stone 1:
x = ut - (g/2)t^2 = t(u - gt/2)

For stone 2:
x = u(t + T) - (g/2)(t^2 + 2tT + T^2)

.

I don't understand the use of (t + T ) for the second stone.

Reply 8

username9816

14

Depends how you look it at it. I looked at it as; t + T, as theres the time in air the along with waiting T seconds to get going. So Total Time = t + T. It's just that for x > 0, t > 0.

Reply 9

steve2005

17

Nima

Depends how you look it at it. I looked at it as; t + T, as theres the time in air the along with waiting T seconds to get going. So Total Time = t + T. It's just that for x > 0, t > 0.

I see.

Thanks

oh no, mechanics.

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