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iragen [17]
3 years ago
9

A brick is dropped from rest from a height of 4.9 m. how long does it take the brick to reach the ground?

Physics
2 answers:
Igoryamba3 years ago
8 0

We know the equation of motion , s =ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, a is the acceleration and t is the time taken.

In this case a brick is dropped from rest from a height of 4.9 m, so displacement, s = 4.9 m

Since the brick is dropped from rest, u = 0 m/s

Acceleration on brick = acceleration due to gravity = 9.81 m/s^2

Substituting

   4.9 = 0*t+\frac{1}{2} *9.81*t^2\\ \\ 4.9 = 4.9t^2\\ \\ t^2=1\\ \\ t =1 second

So the brick will reach ground after 1 second.


wlad13 [49]3 years ago
7 0

The brick takes 1 second to reach the ground

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Height = h = 4.9 m

Gravitational Acceleration = g = 9.8 m/s²

Initial Velocity = u = 0 m/s

<u>Unknown:</u>

time taken to reach the ground = t = ?

<u>Solution:</u>

This problem is about Free Fall.

When an object experiences free fall, the acceleration experienced by the object is the same as the acceleration due to gravity which is equal to 9.8 m/s².

h = ut + \frac{1}{2}gt^2

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

4.9 = 0 + 4.9t^2

t^2 = 1

t = 1 ~ \text{second}

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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