A lift starts from the top of a mine shaft and descends with a constant speed of 10 m/s. 4 s later a boy throws a stone vertically upwards from the top of the shaft with a speed of 30 m/s. Find when and where stone hits the lift [take: g=10 m/s²]

To find when and where the stone hits the lift, we need to determine the time it takes for the stone to reach the lift and the distance traveled by both the stone and the lift during that time.

Let's start by finding the time it takes for the stone to reach the lift.

Given that the stone was thrown upward with an initial velocity of 30 m/s and the acceleration due to gravity is 10 m/s², we can use the kinematic equation:

v = u + at

Where:
v = final velocity (0 m/s as the stone reaches its highest point)
u = initial velocity (30 m/s)
a = acceleration (-10 m/s² due to gravity)
t = time taken

Rearranging the equation, we get:

t = (v - u) / a

Substituting the values, we have:

t = (0 - 30) / (-10)
t = 3 seconds

Therefore, it takes 3 seconds for the stone to reach its highest point.

Now, let's determine the distance traveled by the lift during this time.

Since the lift is descending with a constant speed of 10 m/s, we can use the formula:

distance = speed × time

Substituting the values, we have:

distance = 10 m/s × 3 s
distance = 30 meters

Therefore, the lift travels a distance of 30 meters during the 3 seconds.

To find where the stone hits the lift, we can calculate the distance traveled by the stone during the 3 seconds.

Using the formula for vertical motion:

distance = initial velocity × time + (1/2) × acceleration × time²

Substituting the values, we have:

distance = 30 m/s × 3 s + (1/2) × (-10 m/s²) × (3 s)²
distance = 90 m + (1/2) × (-10 m/s²) × 9 s²
distance = 90 m - 45 m
distance = 45 meters

Therefore, the stone hits the lift 45 meters above the initial position of the lift.

In conclusion, the stone hits the lift 3 seconds after being thrown and 45 meters above the initial position of the lift.

129

For the stone to hit the lift, displacement of both the lift and the stone must be equal.


Let's say it hits in t second
30t-0.5gt²=-10t
Solve for t and it's the answer

48

40

let hi be top of mine shaft

position of lift= -10t
position of stone=30t-4.9(t-4)^2

set these equal, solve for t.

then, solve for lift final postion at t.