# Answer to Question #12829 in Mechanics | Relativity for Andrei

Question #12829

A ball is thrown horizontally from a height of 5.5meters with an initial speed of 25m/sec.What will be the magnitude of its velocity.?

Expert's answer

V = sqrt(25^2 + g^2 * t^2), where t = (0, tmax).

h = g * tmax^2 / 2

tmax =

sqrt(2 * h / g) = sqrt(2 * 5.5 / 9.8) = 1.06 s.

V0 = sqrt(25^2 + g^2 * 0) =

25.00

Vmax = sqrt(25^2 + g^2 * tmax^2) = 27.07

So, the magnitude of its

velocity will change from 25.00 m/s to 27.07 m/s.

h = g * tmax^2 / 2

tmax =

sqrt(2 * h / g) = sqrt(2 * 5.5 / 9.8) = 1.06 s.

V0 = sqrt(25^2 + g^2 * 0) =

25.00

Vmax = sqrt(25^2 + g^2 * tmax^2) = 27.07

So, the magnitude of its

velocity will change from 25.00 m/s to 27.07 m/s.

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