Answer to Question #12829 in Mechanics | Relativity for Andrei

Question

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.?

Answer 1

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.