Question

The ball is dropped from the height of 19.6m if mass of ball is one kilogram then gain its KE can be when it hits the surface of earth.​

Answer 1

Answer:

gain in the Kinetic energy = 192.08 joule

Step by step explanations :

given that,

The ball is dropped from the height of 19.6m

here,

initial velocity of the ball = 0 m/s [ It was at rest]

height from which ball was dropped

= 19.6 m

gravitational acceleration(g) = - 9.8 m/s²

let the velocity of the ball when it will hit the ground be v

now,

we have,

initial velocity(u) = 0 m/s

final velocity = v

gravitational acceleration(g) = -9.8 m/s²

height(h) = 19.6m

by the gravitational equation of motion,

v² = u² + 2gh

putting the values,

v² = (0)² + 2(9.8)(19.6)

v² = (19.6)(19.6)

v² = 19.6²

v = 19.6 m/s

so,

velocity at which it will hit the ground

= 19.6 m

now,

we know that,

Kinetic energy = ½ mv²

where,

m = mass the object

given mass of the ball = 1 kg

v = velocity = 19.6 m/s

so,

putting the values,

= ½ × 1kg × 19.6 m/s × 19.6 m/s

= 9.8 × 19.6

= 192.08 kgm/s²

= 192.08 joule

so,

gain in the Kinetic energy

= 192.08 joule

Answer 2

Answer:

Kinetic energy of ball = 192.08 m

Step by step explanations :

given that,

when the ball was dropped from a height :

mass(m) = 1 kg

initial velocity(u) = 0 m/s

final velocity(v) = v

gravitational acceleration(g) = -9.8 m/s²

[because g is always negative]

height(h) = 19.6m

v² = u² + 2gh

v² = 0² + 2(9.8)(19.6)

v² = 384.16

v = √384.16

v = 19.6 m/s

so,

final velocity of the ball = 19.6 m

Kinetic energy = ½ mv²

= ½ × 1 × 19.6²

= ½ × 19.6 × 19.6

= 192.08,

Kinetic energy of the ball by hitting ground = 192.08 meter