An object is thrown straight upward from the edge of a building with a velocity of 20 ms-1. Where will the object be 5 s after it was thrown?

Answer 1

22.5 m above the point from which it was thrown

Answer 2

h = Hi + Vi t - 4.9 t^2
h = Hi + 20 * 5 - 4.9 * 25
h = Hi + 100 - 122.5
= Top of building - 22.5 meters

Answer 3

So below, not above