Question

From the top of a tall building a stone is just dropped. Below the top at a distance of 1.25 m, there is a window whose height is 3.75 m. If acceleration due to gravity is 10 m/s2, the time taken by the stone to travel from the top to the bottom of the window is 1/4S​

Answer 1

Answer:

1/4 S​

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Answer 2

Correct Question:

From the top of building, a stone is just dropped. Below at a distance of $1.25 m$, there is a window with height, $3.75 m$. If acceleration (due to gravity) is $10 m/s^2$, the time taken by the stone to cover the distance from the top of building to the bottom of the window is ?

Answer:

The time taken by stone to travel the required distance is found to be 0.5 seconds.

Explanation:

The Newton's equation that will be used in this kind of question where the distance and acceleration is given, and the time to cover the given distance is to be calculated, is given by:

$s=ut+\frac{1}{2}at^2$

where 's' is the distance, 'u' is the initial velocity, 't' is the time taken and 'a' is the acceleration of the motion.

Now, we are given that the distance of the window from the top of the building to the window is 1.25 m.

The initial velocity of the stone will be zero as the stone was at rest at the start.

Thus, substituting all values, we get:

$1.25=(0)t+\frac{1}{2}(10)t^2$

Simplifying this, we get:

$1.25=5t^2$

Separating the variables, we get:

$0.25=t^2$

Taking square roots both sides, we get:

$t=\sqrt{0.25}$ sec

t = 0.5 secs.

Thus, the time taken by stone to travel from top to the window is found to be 0.5 seconds.

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