Question

Aball is thrown directly downward with an initial speed of 10 m/s from a height of 40m. When does the ball strike the ground

Answer 1

$S = ut + \frac{1}{2} a {t}^{2} \\ \\ H = ut + \frac{1}{2} g{t}^{2} \\ \\ 40 = 10t + 5{t}^{2} \\ \\ {t}^{2} + 2t - 8 = 0 \\ \\ t = \frac{- 2 \pm \sqrt{4 - 4 \times 1 \times (-8) }}{2 \times 1} \\ \\ t = \frac{-2 \pm \sqrt{36}}{2} \\ \\ t = - 1 \pm 3 \\ \\ t = 2 \: secs \: \: \: or \: \: \: t = -4 \: secs \\ \\ Since, \: t \: cannot \: be \: negative \: so \: t \ne - 4 \: secs \\ \\ So \: the \: answer \: is \: \huge{\boxed{t = 2 secs}}$