Question

The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone's velocity just before it strikes the ground?

Answer 1

use principle of conservation of energy.
gravitational potential energy at the top of the building = kinetic energy just before it strikes the ground
=> mgh = 1/2 mv^2
the only unknown is v.

Answer 2

h = 1/2 a t^2

Answer 3

find t

Answer 4

and subs in the eq V = u + at

Answer 5

u =0 becoz the stone is dropped

Answer 6

@Vaidehi09 what is m and v?

Answer 7

assume m to be the mass of the body.
and v is its velocity just before it strikes the ground.

Answer 8

lol don't confuse that guy

Answer 9

@DLS lol i know right!

Answer 10

guys...using conservation is the easiest approach here!

Answer 11

you don't have the required data to apply convo

Answer 12

after simplification, u'll have
gh = 1/2 v^2
u know g. u know h. find v.
who says we don't have the req data?

Answer 13

sorry i didn't thought that mass would eliminate

Answer 14

-_-

Answer 15

You can use energy or kinematics to solve the problem
With energy is mgh=mv^2/2 where m-mass of the body ,v-velocity ,h is the height
m>0 then you can eliminate it then v^2=2gh
v= \[\sqrt{2gh}\]

Answer 16

square root (2gh) so 2 x 318 x 9.81 =6239.16. Take square root gives 78.988 m/s.
The derivation is from 1/2 mv^2 = mgh
so mv^2 = 2mgh
then v^2 = 2gh
so v = SqRoot (2gh)