Question

a ball is thrown upward with a speed of 120 m/s. its speed at mid point ? take g= 10 m / s2

Answer 1

$h = \frac{ {u}^{2} }{2g} \\ \frac{h}{2} = \frac{ {u}^{2} }{4g} \\ \frac{h}{2} = \frac{120 \times 120}{4 \times 10} \\ \frac{h}{2} = 360 \: m$
$x = ut - \frac{1}{2} g {t}^{2} \\ 360 = 120t - 5 {t}^{2} \\ {t}^{2} - 24t + 72 = 0 \\ t = \frac{ - b + - \sqrt{ {b}^{2} - 4ac}}{2a} \\ t = \frac{24 + - \sqrt{576 - 288} }{2} \\ t = \frac{24 + - \sqrt{288} }{2} \\ t = \frac{12(2 + - \sqrt{2}) }{2} \\ t = {12 + 6 \sqrt{2} } \: or \: 12 - 6 \sqrt{2}$
please follow me