Answer to Question #163584 in Mechanics | Relativity for Si

Question

From an elevated point A, a stone is projected vertically upward. When the stone reaches a distance h below A, its velocity is double what it was at height h below A. Show that the greatest height obtained by the stone above A is 5h/3.

Answer 1

Let u be the velocity with which the stone is projected vertically upwards.

$v_{-h}=2V_h \\ v_{-h}^2=4V_h^2 \\ u^2 - 2g - (-h) = 4(u^2 -2gh) \\ u^2 = \frac{10gh}{3} \\ h_{max} = \frac{u^2}{2g}= \frac{5h}{3}$