Question

a hammer of mass 1 kg falls freely from a height of 2 m. calculate

(1) the velocity of hammer
(2) the kinetic energy of the hammer just before it touches the ground.​

Answer 1

Answer:

u = 0

a = 10 m/s

h = 2m

v = under root 2 * h*a

under root 2*2*10

= 2 * under root 10

k. e = mv^2/2

mass = i kg

=v^2 = 40m/s

1*40/2

40 is divided by 2 = 20

1*20 = 20j is k.e.

Explanation:

Answer 2

1. Velocity of hammer, v = 6.26 m/s

2. Kinetic energy, E = 19.59 J

Explanation:

It is given that,

Mass of the hammer, m = 1 kg

Height, h = 2 m

(1) Let v is the velocity of hammer. It can be calculated using conservation of energy as :

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 2}$

v = 6.26 m/s

(2) Let E is the kinetic energy of the hammer just before it touches the ground.​ It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 1\times (6.26)^2$

E = 19.59 J

Learn more,

Conservation of energy

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