Question

a stone is dropped from the edge of a roof find how long does it take to fall 4.9 m and how fast does it move at the end of the Fall what is the acceleration after 1 and after 2 second

Answer 1

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Answer 2

Given:

Height $=4.9m$

Acceleration due to gravity $=9.8m/s^{2}$

To find:

1. Time of fall of the stone $(t)$.
2. Velocity of the fall of stone $(v)$.
3. Acceleration of the stone after 1 and 2 seconds of the fall.

Solution:

Step 1

We have been given that a stone in rest is dropped from a certain height "h" and asked to calculate the time "t" taken,

Hence, we have here is

$s=h=4.9m$   $;u=0m/s$   $;a=g=9.8m/s^{2}$  

Hence,

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

substituting the given values in the equation, we get

$4.9=(0)t+\frac{1}{2}(9.8)t^{2} \\9.8=9.8t^{2}$

Hence,

$t=1s$

Hence, the ball will travel a distance of 4.9 m in 1 s.

Step 2

Now,

We know the ball has traveled 4.9 m in 1 s and hence, all the potential energy that the ball had at 4.9 m will be converted into kinetic energy or given as the velocity of the stone, according to Law of Conservation of energy.

Hence, at the end of the fall, Potential energy of the ball = Kinetic energy

Mathematically,  $mgh=\frac{1}{2}mv^{2}$

From here, we get

$v^{2}=2gh$   $;$ $or$

$v=\sqrt{2gh}$

Substituting the given values in the equation, we get

$v=\sqrt{2(9.8)(4.9)}$   $;$ $or$

$v=9.8m/s$

Step 3

The ball while falling, acts under the same acceleration that is the acceleration due to gravity (g) hence, the acceleration on the ball remains same

$a=9.8m/s^{2}$

Final answer:

Hence,

1. The ball takes $1 s$ in falling a distance $4.9 m$.
2. The ball travels with a speed of $9.8 m/s$ towards the end of fall.
3. The acceleration of the ball is $9.8m/s^{2}$.