Question

A ball of mass 2 kg is project horizontal with a velocity 20m/s from building of height 15m the speed with body hits the ground

Answer 1

Answer:

velocity of ball by hitting ground

= 10√7 m/s

Explanation:

Given that,

mass of the body = 2 kg

velocity by which it was thrown = 20 m/s

height of the building = 15 m

FIRST METHOD :

BY THE NEWTON's GRAVITATION LAW

NOTE : Velocity of hitting ground of any object does not depends upon the angle of throwing and mass of the object.

so,

let the final velocity of the ball be v

now we have ,

Initial velocity(u) = 20

final velocity(v) = v

displacement(h) = 15 m

gravitational acceleration(g) = 10 m/s²

by the gravitational equation of motion

v² = u² + 2gh

putting the values,

v² = 20² + 2(10)(15)

v² = 400 + 300

v² = 700

V = 10√7 m/s

____________

SECOND METHOD

BY THE WORK ENERGY THEOREM

We know that,

Work done = Δ Kinetic energy

so,

mgh = ½ m (v² - u²)

putting the values,

2 × 10 × 15 = ½ × 2 (v² - 20²)

300 = v² - 400

v² = 300 + 400

v² = 700

v = 10√7 m/s

_______________

velocity of ball by hitting ground

= 10√7 m/s

Answer 2

Answer:

Final velocity of the ball = 10√7 m/s

Explanation:

Given,

mass of the body = 2 kg

velocity by which it was thrown = 20 m/s

height of the building = 15 m

by the work energy theorem

energy will be conserved

so,

potential energy at its highest point = Δ kinetic energy of just hitting the ground

so,

mgh = ½ m(v² - u²)

gh = (v² - u²)/2

h = 15 m

u = 20 m/s

so,

10(15) = (v² - 20²)/2

v² - 400 = 300

v² = 700

v = 10√7 m/s

so,

Final velocity of the ball = 10√7 m/s