Question

An object is falling down from a building .If it takes 3 s to reach ground.the distance b/w building and ground is _______ m ?

Answer 1

Question

An object is falling down from a building. If it takes 3 s to reach the ground, the distance between the building and the ground is _______ m.

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Answer

Initial Velocity (u) = 0 m/s

Acceleration (a) = 9.8 m/s² [Since acceleration for any object falling down is 9.8 m/s²]

Time (t) = 3 sec

Distance (s) = ?

For finding the distance covered in this case, we'll apply the 2nd Equation of Motion.

2nd Equation of Motion → $s = ut + \dfrac{1}{2}at^{2}$

We'll substitute the variables from the equation and find the value of 's' now.

$\implies s = 0(3) + \dfrac{1}{2} (9.8)(3)^{2}$

$\implies s =0 + \dfrac{1}{2} (9.8)(9)$

$\implies s = (0.5) (9.8)(9)$

$\implies s = 44.1 \ m$

∴ The distance between the building and the ground is 44.1 m

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Answer 2

QUESTION:-

An object is falling down from a building .If it takes 3 s to reach ground.the distance between  building and ground is _______ m ?

EXPLANATION:-

In the given question-:

• Initial velocity(u)=0 m/s [SINCE THE OBJECT FALL FROM REST]
• time taken(t)=3 s

• Acceleration due to gravity(g)=9.8 m/s²

• Height(H)=??

Now let's use 2nd equation of kinematics

H=ut+1/2×a×t²

Putting values in the formula we get:-

→H=1/2×9.8×(3)²

→H=4.9×9

→H=44.1 m

So the height is 44.1 m

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