Question #40e94

1 Answer
Nov 2, 2016

Answer:

#sf(1960color(white)(x)m)#

Explanation:

I just keep things simple and say that, for an object falling from rest:

#sf(s=1/2"gt"^(2))#

#:.##sf(s=1/2xx9.8xx20^(2)=1960color(white)(x)m)#

It looks like you are over complicating things. There is no need to find the velocity.

It seems that you have found the velocity of the object when it hits the ground but the velocity is not constant during the journey.

Then you have treated velocity as being equal to #sf(at)# in the displacement equation and squared it as in #sf((at))^(2)# when it should be #sf(at^2)# which I think is where you have gone wrong.