Question

from the top of a tall building a stone is just dropped. below the top at a distance of 1.25m, there is a window whose height is 3.75m. if acceleration due to gravity is 10m/s, the time taken by the stone to travel from the top to the bottom of the wndow is

Answer 1

Answer:

9.8 m/s

Explanation:

Here Initial velocity (u) = 0

Final velocity (v) = 58.8 m/s

Time (t) = 6 seconds

We know v=u+at

58.8=a×6

⇒6a=58.8

⇒a=

6

58.8

∴a=9.8m/s

2

∴ The value of acceleration due to gravity = 9.8 m/s

2

Answer 2

Answer:

The time taken = 1 s.

Explanation:

Lets assume the time taken by the stone to fall from top of the building to bottom of the window is 't'.

• At the top of building the initial velocity of stone = u = 0

• From the top of the building to the bottom of window, total distance,

                        s = 1.25 + 3.75 = 5 m

• Acceleration due to gravity, g = 10 m/$s^{2}$
• From the second equation of motion we know that

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

    Putting all the values, 5 = (0 x t) + ($\frac{1}{2}$ x 10 x $t^{2}$)

                                        ⇒ 5 = 0 + 5$t^{2}$

                                        ⇒  $t^{2}$ = 1

                                        ⇒ t = 1

∴ Stone will take 1 second time to cover the distance.