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A body is projected vertically upwards. If $t_1$ and $t_2$ be the instants at which it is at a height $h$ above the point of projection, while ascending and descending respectively.Then find $h$ !!

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written 2.1 years ago by | • modified 2.0 years ago |

- u = velocity of projection (upward).
- s = displacement after time t = h
- a = acceleration (downward) = - g

$Using, s=u t+\frac{1}{2} a t^{2}$ We have,

$h = u t - \frac{1}{2} g t² $

2h = 2ut - gt²

gt² - 2ut + 2h = 0 .

This is an equation in t which shows that for a given h there are two values of t.

Multiplying the above equation by g through out we get

g²t² - 2ugt + 2gh = 0

(gt)² - 2(gt)(u) + u² - u² + 2gh = 0

(gt - u)² = u²- 2gh

$gt - u = ± \sqrt{(u²- 2gh)}$

$gt = u ± \sqrt{(u²- 2gh)}$

$t = \frac{{u ± \sqrt{(u²- 2gh)}}} { g}$

$t1 = \frac{{u + \sqrt{(u²- 2gh)}}} { g}$ and $t2 = \frac{{u - \sqrt{(u²- 2gh)}}} { g}$

Multiplying, $(t1)(t2) = \frac{{u² - (u² - 2gh)}} { g²} = \frac{2gh }{ g²}$

$(t1)(t2) = \frac{2h }{ g}$

$h = \frac{1}{2}g(t1)(t2)$

Therefore the final answer $h = \frac{1}{2}g(t1)(t2)$

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