A coin is released inside a lift at a ht of 2m from the floor of the lift. The ht of the lift is 10 m . the lift is moving with an acceleration of 9m/s² downwards. The time after which the coin will strike with lift is: (g= 10 m/s)

Question

A coin is released inside a lift at a ht of 2m from the floor of the lift. The ht of the lift is 10 m . the lift is moving with an acceleration of 9m/s² downwards. The time after which the coin will strike with lift is: (g= 10 m/s)

👇

2 answers
(click or scroll down)

👇

Answer 1

bobpursley answered
10 years ago

Both have the same initial velocity. The coin has the greater acceleration. The coin has to travel 2 m further than the bottom of the lift.

distancelift=vi+9t^2/2
distancecoin=vi+10t^2/2
or
vi+9t^2/2-2=vi+10t^2/2
.5t^2=2

t=2 seconds check my math.

Answer 2

Explain Bot answered
1 year ago

To find the time after which the coin will strike with the lift, we can use the equation of motion:

h = ut + (1/2)at^2

Where:
h = initial height = 2m
u = initial velocity = 0m/s (since the coin is released, it starts from rest)
a = acceleration = -9m/s² (negative because the lift is moving downwards)
t = time

We know that the final height when the coin strikes the lift is 0m (since it will be on the floor of the lift). Therefore, we can rearrange the equation to solve for time:

0 = (1/2)(-9)t^2 + 0t + 2

Simplifying the equation:

-4.5t^2 + 2 = 0

Dividing both sides of the equation by -4.5:

t^2 = 2/4.5
t^2 = 4/9

Taking the square root of both sides:

t = √(4/9)
t = 2/3

Therefore, the time after which the coin will strike with the lift is 2/3 seconds.