Solve-quadratic-equation-about-t-1-h-v-0-t-1-2-gt-2-2-x-v-0-t-1-2-at-2-solve-v-v-2-v-0-2-2ax-

Question Number 170546 by libaolin last updated on 26/May/22

S o l v e : q u a d r a t i c e q u a t i o n a b o u t t :

1 . h = v_{0} t + \frac{1}{2} {gt}^{2}, 2 . x = v_{0} t + \frac{1}{2} {at}^{2}

solve v : v^{2} - v_{0}^{2} = 2 a x

Answered by Rasheed.Sindhi last updated on 26/May/22

1 . h = v_{0} t + \frac{1}{2} {gt}^{2}

{2 v}_{0} t + {gt}^{2} = 2 h

{gt}^{2} + {2 v}_{0} t - 2 h = 0

t = \frac{- {2 v}_{0} \pm \sqrt{{({2 v}_{0})}^{2} - 4 (g) (- 2 h)}}{2 g}

= \frac{- {2 v}_{0} \pm \sqrt{{4 v}_{0}^{2} + 8 gh}}{2 g}

= \frac{- {2 v}_{0} \pm 2 \sqrt{v_{0}^{2} + 2 gh}}{2 g}

= \frac{- v_{0} \pm \sqrt{v_{0}^{2} + 2 gh}}{g}

2 . x = v_{0} t + \frac{1}{2} {at}^{2}

Similarly,

t = \frac{- v_{0} \pm \sqrt{v_{0}^{2} + 2 ax}}{a}

3 . v^{2} - v_{0}^{2} = 2 a x

v^{2} = v_{0}^{2} + 2 a x

v = \pm \sqrt{v_{0}^{2} + 2 a x}

Commented by peter frank last updated on 26/May/22

thank you