If you toss a ball upward with a certain initial speed, it falls freely and reaches a maximum height h. By what factor must you increase the initial speed of the ball for it to reach a maximum height 3h?

Let [tex]g[/tex] denote the gravitational field strength ([tex]g \approx 9.8\; {\rm m\cdot s^{-2}}[/tex] near the surface of the earth.) If the drag on the ball is negligible, the ball will be constantly accelerating downward at [tex]g[/tex] while in the air.

Let [tex]u[/tex] denote the initial velocity of this ball. In this example, [tex]u\![/tex] will be the velocity at which the ball is tossed upwards.

When the ball is at maximum height, its velocity will be [tex]0[/tex]. Let [tex]v[/tex] denote this velocity.

Let [tex]x[/tex] denote the displacement of this ball when it reached maximum height relative to when the ball was initially tossed upward. In this example, [tex]x = h[/tex].

Let [tex]a[/tex] denote the acceleration of this ball. Under the assumptions, [tex]a = g \approx (-9.8)\; {\rm m\cdot s^{-2}}[/tex].

The SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] relates these quantities.

Note that since [tex]v = 0[/tex], [tex]x = h[/tex], and [tex]a = (-g)[/tex], this equation becomes:

[tex]0^{2} -u^{2} = 2\, (-g)\, h[/tex].

[tex]u^{2} = 2\, g\, h[/tex].

[tex]u = \sqrt{2\, g\, h} = (\sqrt{2\, g})\, (\sqrt{h})[/tex].

Therefore, replacing [tex]h[/tex] with [tex]3\, h[/tex] will only increase the initial velocity of the ball [tex]u[/tex] by a factor of [tex]\sqrt{3}[/tex].