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September 12th, 2016, 11:01 PM   #1
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Velocity and time

A particle is projected vertically upwards with velocity u m/s. Its height is h after t1 and t2 seconds. Prove that t1 × t2 = 2h/g.

Last edited by skipjack; September 13th, 2016 at 02:33 AM.
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September 12th, 2016, 11:32 PM   #2
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Set up the quadratic equation describing the motion at height h.

You know that t1, t2 are roots of this quadratic equation.

Using the general quadratic formula, find out the expression for the product of two roots of a quadratic equation. These will be in terms of the $a,b,c$ in the quadratic formula.

Using your equation you set up in step 1, note the values of these coefficients and plug them into the result you got above.

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$h(t) = -\dfrac 1 2 g t^2 + u t$

solve for $t_1,~t_2$

$h = -\dfrac {g t^2}{2}+u t$

$-\dfrac {g t^2}{2}+u t-h=0$

you'll get two roots in the form of $t_1,t_2 = \dfrac{-b \pm \sqrt{b^2 - 4 a c}}{2a}$

where here $a=-\dfrac{g}{2},~b=u,~c=-h$

Now if you you multiply those two roots out you'll find that

$t_1 \cdot t_2 = \dfrac c a = \dfrac{-h}{-g/2} = \dfrac {2h}{g}$

Last edited by skipjack; September 13th, 2016 at 02:34 AM.
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September 13th, 2016, 08:23 AM   #3
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