Question

A ball dropped from the top of the building reaches the ground in 6 seconds. Find the velocity with which it will hit the ground if value of g is 9.8 ms-1​

Answer 1

Answer:

In the present event a ball is being dropped (the term means that initial velocity=0)

from a height h

h = 19.6 m = 2 times the numerical value of g, the acceleration due to gravity

we know how accelerations operate ..it will add per sec the value of g to the velocity

v = g.t

but the time taken to cover h =19.6 m ; using h = (1/2) g . t^2

t = sqrt ( 2.h /g) = 2

so the velocity of striking the ground v = 9.8 m/s^2 x 2 s = 19.6 m/

Explanation:

Answer 2

Answer:

58.5 m/s

Explanation:

From 1st Law of Newton,

v = u + at

= 0 + 9.8 × 6 ( u = 0, since ball is dropped)

= 9.8 × 6

= 58.5 m/s

HOPE THIS HELPS YOU......