Question

Necesito ayuda para resolver esto porfa

Answer 1

Epg = Ek

$mgh = \frac{1}{2} m {v}^{2} \\ \\ 2mgh = m {v}^{2} \\ \\ 2gh = \frac{m {v}^{2} }{m} \\ \\ 2gh = {v}^{2} \\ \\ h = \frac{ {v}^{2} }{2g}$
$\\ h = \frac{ {30.4}^{2} }{2(9.8)} \\ \\ h = \frac{924.16}{19.6} \\ \\ h = 47.15$
La altura es de 47, 15 [m]