Question

A stone is is thrown vertically upward from the top of a tower with a velocity u and it reaches the ground with a velocity 3u the height of the tower is
(A) 3u²/g
(B) 4u²/g
(C) 6u²/g
(D) 9u²/g​

Answer 1

Answer:

(B) 4u²/g

Explanation:

given that,

A stone is is thrown vertically upward

from the top of a tower with a velocity u

here,

initial velocity of the stone = u

also, given that,

it reaches the ground with a velocity 3u

so,

now,

we have,

initial velocity = u

final velocity = 3u

gravitational acceleration(g) = -10 m/s²

now,

by the gravitational equation of motion

v² = u² + 2as

and here,

initial velocity when the stone passes the initial position after throwing will be equal to the

velocity at which it was thrown

so,

here,

we can say that,

initial velocity from the tower to the ground = u

so,

now,

we have,

initial velocity = u

final velocity = 3u

so,

by the gravitational equation of motion

v² = u² + 2gh

here,

h is the height of the tower

now,

putting the values,

(3u)² = u² + 2gh

9u² = u² + 2gh

2gh = 9u² - u²

2gh = 8u²

h = 8u²/2g

h = 4u²/g

so,

Height of the tower = 4u²/g

___________________

CORRECT OPTION:

(B) 4u²/g

Answer 2

Answer:

(B) 4u²/g

Explanation:

Initial velocity = u

final velocity = 3u

v² = u² + 2gh

(3u)² = u² + 2gh

9u² - u² = 2gh

h = 8u²/2g

h = 4u²/g