Question

An object is thrown directly downwards from a height of 60m with an initial velocity of 11m/s. What will be its velocity on impact?

v^2 = U^2 + 2as
a= 9.81m/s^2
u = 11m/s
s = 60m

v^2 = 11m/s + 2 * 9.81m/s^2 * 60
= 1188.2m^2/s^2
= square root of 1188.2m^2/s^2
= 34.47m/s

Is this done correctly??

Answer 1

v^2 = 11m/s + 2 * 9.81m/s^2 * 60
should read:
v =sqrt( 11m/s ² + 2 * 9.81m/s^2 * 60 )

I get about 36 m/s