Question

A ball of mass 1 kg is dropped from a height of 5 metre a) find the kinetic energy of the ball just before it reaches the ground b) what is its speed at this instant?

Answer 1

Hey Mate

your answer is here

let potential energy is zero at ground

then

potential energy at height 5m = mg = 1×10 = 10J

so ,

kinetic energy at ground = 10J

according to energy conversation law

so,

1/2mv^2 = 10

v^2 = 20

v = root 20 m/s

Answer 2

Answer:

my answer is correct answer see here .

^ 2 this means root 2 ok.

Explanation:

(A) mass = 1 kg

height = 5m

g = 10 m/s ^2

initial kinetic energy (K.E) =0

vpotential energy (P.E ) = mgh

=1 × 10 × 5

= 50 joules .

P.E + initial K.E = final KE

=50 + 0 =50 .

k.E = 50 joule

(B)

K.E = 1 / 2 mv^2 = 50 joule .

= 1/2 × 1 × v^2 = 50.

v^2 = 50 × 2

v^2 = 100

v^2 = _/100

v = 10 m/s ^2

speed = 10 m/s ^2

this is the correct ans .you can check them by your teachers also.

this is correct .

and then please like and rate and follow also .

by guys