Question

11. A ball thrown vertically upwards from level ground is observed twice, at a height H above the ground within
interval t. The initial velocity of the ball was​

Answer 1

Answer:

√(25t^2+20H)

Let initial velocity was u . At time t it's height can be given by

$s = ut - \frac{1}{2} g {t}^{2} \\ s = ut - 5 {t}^{2} \\ 5 {t}^{2} - ut + s = 0$

when s = h

$5 {t}^{2} - ut + h = 0$

Let two roots of this quadratic are a and b which are times when the ball is on height h .

From quadratic we can write

$a + b = \frac{u}{5} \: \: and \: \: ab = \frac{h}{5}$

Given interval t is nothing but a-b

$(a - b) {}^{2} = {(a + b) {}^{2} - 4ab } \\ \\ {t}^{2} = ( \frac{u}{5}) {}^{2} - 4( \frac{h}{5} ) \\ {u}^{2} = 25 {t}^{2} + 20h \\ u = \sqrt{25 {t}^{2} + 20h }$