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/s2] (Give value of 20/6 = 49)
Answers
Answer:
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Answer:
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.