Question

a brick is dropped from rest from a heightof 5.0m.how long does it take for the brick to reach the ground?​

Answer 1

Given:

• Initial velocity (u) = 0 m/s
• Height (h) from the ground = 5 m

To find:

• Time taken by the brick to reach the ground.

Solution:

☆ Acceleration is defined as the rate of change of velocity.

We know that,

$\boxed {\mathfrak {\pink {s = ut + \dfrac {1}{2} at^{2}}}}$

Where,

s = distance

u = Initial velocity

t = time

a = acceleration

But, we are given the height instead of the distance. So, we can rewrite the equation as,

$\boxed {\mathfrak {\blue {h = ut + \dfrac {1}{2} gt^{2}}}}$

Where,

h = height

g = acceleration due to gravity

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

Now, substitute the values,

5 = 0×t + ½ × 9.8t²

5 = ½ × 9.8t²

5 = 4.9t²

5÷4.9 = t²

1.02 = t²

$\sf \sqrt{1.02} = t$

$\boxed {\bf {\orange {1s = t}}}$ ---> Approx

Time taken:

The brick will take 1 second to reach the ground

Answer 2

Initial velocity = 0 m/s (rest)

Initial height = 5 m

Acceleration = g = + 10 m/s² (up)

We know,

$\boxed{\bf{\Delta h = ut + \frac{1}{2} gt^2}}$

⇒ $\bf{h = \frac{1}{2} gt^2}$

⇒ $\bf{ t^2 = \frac{2h}{g} }$

⇒ $\bf{t^2 = \frac{2 \times 5 \ m}{10 \ ms^{-2}}}$

⇒ $\bf{t^2 = 1 \ s^2}$

⇒ $\bf{ t = \sqrt{1 \ s^2} = 1 \ s}$.

∴ The brick will take about 1 second to reach the surface of the Earth.