A lift starts from the top of a mine shaft and descends with a constant speed of $10 m / s$ . $4$ sec 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 $\frac{x}{3}$ (in m)
[Take: $g = 10 m / s^{2}$ ] (given value of $20 \sqrt{6} = 49$ )

Open in App

Solution

Let it meets it at time "t"
$X = (t + 4) 10$ ...........(1)
$F o r b a l l - ------- - :$
$- x, u = 30 m / s, u = - g, t$
$- X = 30 t - \frac{10}{2} t^{2}$
$- (t + 4) 10 = 30 t - 5 t^{2}$
$- 10 t - 40 = 30 t - 5 t^{2}$
$5 t^{2} - 40 t - 40 = 0$
$t^{2} - 8 t - 8 = 0$
$t = \frac{8 \pm \sqrt{64 + 32}}{2}$
$t \Rightarrow \frac{8 + \sqrt{96}}{2}$
$t \Rightarrow 4 + \sqrt{24}$
Put the value of t in (1)
$X = (4 + \sqrt{24}) 10$
futher solving we get,
$X = 30$

Suggest Corrections

Similar questions

An open lift is coming down from the top of a building at a constant speed v=10 m/s. A boy standing on the lift throw a stone vertically upwards at a speed of 30 m/s w.r.t. himself. The time after which he will catch the stone is:

Q. An open lift is coming down from the top of a building at a constant speed

v = 10 m / s

. A boy standing on the lift throws a stone vertically upwards at a speed

30 m / s

w.r.t. himself. The time after which he will catch the stone is:

Q. A lift is coming down with a downward acceleration of 2

m / s^{2}

. A boy standing in the lift thrown a stone vertically upwards with speed 5 m/s w.r.t. himself, the time after which he will catch the stone is

{

g = 10 m / s^{2}

}

Q. A boy is standing on top of a tower of height 85m and throws a ball in the vertically upward direction with a certain speed. If 5.25 seconds later he hears the ball hitting the ground, then the speed with which the boy threw the ball is (take g = 10 m/s2 , speed of sound in air = 340 m/s)

Q. From the top of a building of height

40 m

, a boy throws a stone vertically upwards with an initial velocity of

10 m s^{- 1}

such that it eventually falls to the ground. At what time will the stone reach the ground?

(g = 10 m s^{- 2})

Join BYJU'S Learning Program

Let it meets it at time "t"X=(t+4)10 ...........(1) For ball–––––––––:−x,u=30m/s,u=−g,t−X=30t−102t2−(t+4)10=30t−5t2−10t−40=30t−5t25t2−40t−40=0t2−8t−8=0t=8±√64+322t⇒8+√962t⇒4+√24Put the value of t in (1)X=(4+√24)10futher solving we get,X=30