My Math Forum Velocity and time

 Applied Math Applied Math Forum

 September 12th, 2016, 11:01 PM #1 Member   Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 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.
 September 12th, 2016, 11:32 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,495 Thanks: 1370 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. profit $h(t) = -\dfrac 1 2 g t^2 + u t$ 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.
 September 13th, 2016, 08:23 AM #3 Member   Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 Thanks

 Tags time, velocity

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post themagicstick1 Physics 2 September 14th, 2015 02:01 AM kunz398 Calculus 2 May 14th, 2015 07:20 AM quarkz Calculus 0 April 18th, 2014 05:34 AM Chikis Algebra 7 October 5th, 2012 10:03 PM condemath Calculus 1 October 25th, 2011 02:53 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top