Question

A heavy object is thrown vertically from the top of a 1.00×10² m building at a Velocity after 10m/s.What is the speed as it reaches the ground (g=9.812m/s²)​

Answer 1

u = 10 m/s

g = 9.812 m/s^2

h = s = 10^2 = 100 m

By the third equation of motion:

v^2 - u^2 = 2gs

=> v^2 - (10)^2 = 2(9.812)(100)

=> v^2 - (10)^2 = 2(9.812)(100)=> v^2 = 1962.4 + 100

=> v = √2062.4

=> v = 45.41 m/s