Question

a ball is dropped from the top of a tall building if it takes 2.8 second to reach the ground. Find the height?​

Answer 1

Explanation:

Given,

Time T = 2.8 s

initial velocity u = 0

find h =?

acceleration due to gravity g = 9.8

S = ut + 1/2at^2

h = 0 + 1/2*9.8*(2.8)^2

h = 4.9*2.8*2.8 = 38.42 m

height is 38.42 m

if u like plz give me brainliast

Answer 2

The three equations of motion are valid for a body with constant acceleration. Here acceleration is due to the gravity of the earth (g) which can be taken as constant for heights that are very small than the radius of the earth. Hence, we can apply the equations of motion here.

The second Equation of Motion states that :

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

          where,

                     S = displacement of body

                     u = initial velocity of body

                     t = time taken

                     a = acceleration of the body

Given , t = 2.8 s

• since motion is due to gravity , a = g = 9.8 m/s²

• since ball is dropped from the top , initial velocity u = 0 m/s

Subsituting these values in 2nd equation of motion we get

                  $S = 0*2.8 + \frac{1}{2}*9.8*(2.8)^{2}$

                      $= 38.416 m$

⇒   The height of the building is 38.416 m