Question

Jun throws a stone straight upward alongside a tree with an initial velocity of 32 ft/sec. The stone rises until it is even with the top of the tree and then falls back to the ground. How tall is the tree?

Answer 1

$\text{Given u = 32ft/sec, let g = 32.174 05ft/}s^2, \text{the formula to reach the maximum height is } \\v^2= u^2 - 2gh \text{ where v =0, since the velocity at maximum height is 0}\\\text{this implies that h = }\frac{u^2}{2g}\\\text{therefore h}=\frac{1024}{ 64.3481} \text{, h = 15.9134ft}$