Question

A lift starts from the top of a mine shaft and descends with a constant speed of 10 m/s . 4s later a boy throws a stone vertically upward from the top of the shaft with a speed of 30m/d .if the stone hits the lift at a distance x below the shaft write the value of x/3

Answer 1

Hello Dear.

Given ⇒
Speed of the lift = 10 m/s.

Distance traveled by the lift in 4 seconds = Speed of the lift × 4
= 10 × 4
= 40 m/s.

Now, Initial Speed of the Stone(u) = 30 m/s.

Final Velocity of the stone(v) = 0

Acceleration = -g
[∵ the Stone is thrown against the gravitational force]
= -10 m/s²


Using the Equation of Motion,

v² - u² = 2aS
⇒ (0)² - (30)² = 2(-10)(S)
⇒ -20 × S = -900
⇒ S = 45 m.

Thus, Distance covered by the Stone in the certain time is 45 m.

∵ v - u = at
∴ (0) - (30) = (-10)t
 10t = 30
⇒ t = 3 seconds.

Thus, time taken by the Stone to reach the height of 45 m is 3 seconds.
∴ Distance travelled by the lift in 3 seconds = 3 × 10
= 30 m.

After Reaching the Height of 45 m, stone starts falling from there and covered the distance of (45 + x) m.

During the Fall,
Initial Velocity(u) = 0.
Acceleration = g.
= 10 m/s².

Time taken by the stone to covers the distance (45 + x) m during the fall = Time taken by the stone to travels the distance of x m.

Using the Second Equation of the motion,
S = ut + 1/2 at²

On Solving the Equating, the Value of x will be 129 m.

Now, x/3 = 129/3
  = 33 m.


Hope it helps.

Answer 2

Answer:

Explanation:

Hey mates !!!!!

Given ⇒

Speed of the lift = 10 m/s.

Distance traveled by the lift in 4 seconds = Speed of the lift × 4

= 10 × 4

= 40 m/s.

Now, Initial Speed of the Stone(u) = 30 m/s.

Final Velocity of the stone(v) = 0

Acceleration = -g

[∵ the Stone is thrown against the gravitational force]

= -10 m/s²

Using the Equation of Motion,

v² - u² = 2aS

⇒ (0)² - (30)² = 2(-10)(S)

⇒ -20 × S = -900

⇒ S = 45 m.

Thus, Distance covered by the Stone in the certain time is 45 m.

∵ v - u = at

∴ (0) - (30) = (-10)t

 10t = 30

⇒ t = 3 seconds.

Thus, time taken by the Stone to reach the height of 45 m is 3 seconds.

∴ Distance travelled by the lift in 3 seconds = 3 × 10

= 30 m.

After Reaching the Height of 45 m, stone starts falling from there and covered the distance of (45 + x) m.

During the Fall,

Initial Velocity(u) = 0.

Acceleration = g.

= 10 m/s².

Time taken by the stone to covers the distance (45 + x) m during the fall = Time taken by the stone to travels the distance of x m.

Using the Second Equation of the motion,

S = ut + 1/2 at²

On Solving the Equating, the Value of x will be 129 m.

Now, x/3 = 129/3

  = 43 m