Answer to Question #270561 in Mechanics | Relativity for Klara

Question

You're standing on the edge of a vertical cliff of height H. The ground below is horizontal. You throw a baseball with speed v in horizontal forward direction. The ball hits the ground below.

Next, your friend throws a second (identical) baseball with the same speed v in forward direction but under an angle π/4 above the horizontal.

The acceleration of gravity is g (pointing downward) and air friction can be ignored.

Derive an expression of the speed v both baseballs should have to touch te ground at exactly the same spot. Express v in the given quantities and constants.

Answer 1

$x=v\cdot t=v\cdot\sqrt{2H/g}$

$x=v\cdot\cos45°\cdot t_0=v\cdot\cos45°\cdot (\sqrt{\frac{2}{g}(\frac{v^2\sin^245°}{2g}+H)}+\frac{v\sin45°}{g})$

$v\cdot\cos45°\cdot (\sqrt{\frac{2}{g}(\frac{v^2\sin^245°}{2g}+H)}+\frac{v\sin45°}{g})=v\cdot\sqrt{2H/g}\to$

$v=\sqrt{Hg/2}$ . Answer