Question

a stone is dropped from the top of a tower of height 1.98 . calculate its final velocity before touching the ground .

Answer 1

Hi friend 

Here,

Height (h) = 1.98 m 

We know ,

v² - u² = 2as ( second equation of motion) 

but in this situation s = h 

and a = acceleration due to gravity (g) 

v² - u² = 2gh

as the stone starts from rest so u  = 0 

now,

v² = 2gh 

v  = √2gh 

This is the formula for finding velocity of any object before touching the ground if we know the height 

so , 

v = √ 2 x 9.8 x 1.98

v = 38.808 m/s 

Answer 2

V^2 = U^2 + 2as
U= 0

Therefore,
V^2 = 2×9.8×1.98
V = 6.229 m/s^2