Question

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. If stone hits the lift at
a distance x below the shaft write the value of x(in m) [Take: g = 10 m/s) (Give value of 20/6 = 49)​

Answer 1

Hey Dear,

◆ Answer -

x = 129 m

● Explaination -

When a boy throws stone upwards, time taken by the stone to reach top is t.

v = u + at

0 = 30 + (-10)t'

t' = 30/10

t' = 3 s

Same time will be taken by stone to return at height of mineshaft.

So total time taken by stone to return to mineshaft after lift start is 4 + 3 + 3 = 10 s.

Now, stone going downwards after that -

s = ut + 1/2 at^2

x = 30 × t + 1/2 × 10 × t^2

x = 30t + 5t^2 ...(1)

For the lift, this eqn will be -

s = u × (t+10)

x = 10 × (t + 10)

x = 10t + 100 ...(2)

Equating (1) & (2),

30t + 5t^2 = 10t + 100

6t + t^2 = 2t + 20

t^2 + 4t - 20 = 0

Solving this quadratic eqn,

t = 2.9 s

Putting this in (2),

x = 10t + 100

x = 10 × 2.9 + 100

x = 129 m

Therefore, stone will hit the lift at 129 m below the shaft.