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A ball is in free fall after being dropped. What willthe speed of the ball be after 2 seconds of free fall?

Sagot :

MDKPfromIDN

So, the speed of the ball after 2 seconds after free fall is 20 m/s.

Introduction

Hi ! I'm Deva from Brainly Indonesia. In this material, we can call this event "Free Fall Motion". There are two conditions for free fall motion, namely falling (from top to bottom) and free (without initial velocity). Because the question only asks for the final velocity of the ball, in fact, we may use the formula for the relationship between acceleration and change in velocity and time. In general, this relationship can be expressed in the following equation :

[tex] \boxed{\sf{\bold{a = \frac{v_2 - v_1}{t}}}} [/tex]

With the following conditions :

  • a = acceleration (m/s²)
  • [tex] \sf{v_2} [/tex] = speed after some time (m/s)
  • [tex] \sf{v_1} [/tex] = initial speed (m/s)
  • t = interval of time (s)

Problem Solving

We know that :

  • a = acceleration = 9,8 m/s² >> because the acceleration of a falling object is following the acceleration of gravity (g).
  • [tex] \sf{v_1} [/tex] = initial speed = 0 m/s >> the keyword is free fall
  • t = interval of time = 2 s

What was asked :

  • [tex] \sf{v_2} [/tex] = speed after some time = ... m/s

Step by step :

[tex] \sf{a = \frac{v_2 - v_1}{t}} [/tex]

[tex] \sf{(a \times t) + v_1 = v_2} [/tex]

[tex] \sf{(10 \times 2) + 0 = v_2} [/tex]

[tex] \boxed{\sf{v_2 = 20 \: m/s}} [/tex]

So, the speed of the ball after 2 seconds after free fall is 20 m/s.

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