Question

a stone is dropped from a helicopter moving in upward direction at 500 m height . it reaches the Earth after 6 seconds. what was the velocity of the helicopter when the stone was dropped.​

Answer 1

Answer:

velocity of the helicopter was dropped.

= 113.3 m/s

Step by step explanations :

given that,

a stone is dropped from a helicopter moving in upward direction at 500 m height

here,

height from which stone was dropped

= 500 meter

also,

it reaches the Earth after 6 seconds.

so,

time taken by the ball to reach ground

= 6 seconds

let the velocity of helicopter when stone was dropped be u

now,

we have,

initial velocity(u) = u

height(h) = 500 m

time taken(t) = 6 seconds

gravitational acceleration(g) = -10m/s²

by the gravitational equation of motion,

h = ut + ½ gt²

putting the values,

500 = u(6) = ½ × (-10)(6)(6)

500 = 6u - 180

6u = 500 + 180

6u = 680

u = 680/6

u = 113.3 m/s

so,

velocity of the helicopter when the stone was dropped.

= 113.3 m/s

Answer 2

Answer:

Explanation:

Given :-

Initial velocity of the helocopter, (u) = u

Height from which stone was dropped, (h)  = 500 meter

Time taken by stone to reach earth, (t) = 6 seconds

Gravitational acceleration of the stone, (g) = 10 m/s²

To Find :-

Velocity of the helicopter when the stone was dropped.

Formula to be used :-

Gravitational equation of motion i.e h = ut + 1/2 gt²

Solution :-

Let the velocity of the helicopter when the stone was dropped is u.

Putting all the valuse, we get

h = ut +1/2 gt²

⇒ 500 = u(6) =1/2 × (-10)(6)(6)

⇒ 500 = 6u - 180

⇒ 6u = 500 + 180

⇒ 6u = 680

⇒ u = 680/6

⇒ u = 113.3 m/s

Hence, the velocity of the helicopter when the stone was dropped is 113.3 m/s.