Question

A ball is dropped form a tall building. It reaches the ground in 3.2 seconds. What is the height of

the building? (g = 10 m.s-2)​

Answer 1

by equation

s = ut +1/2 gt sq.

here:s=?

u=0

t=3.2

g=10

hence, s=0(3.2)+1/2 x 10 x 3.2 x 3.2

s= 5 x 3.2 x 3.2= 51.2

thus s = 51.2 meters

I guess this is the answer

Answer 2

Hi!! Here is the answer:

Since it is a freely falling body, it is a case of uniformly accelerated motion, i.e. the 3 equations if motion can be applied.

Using the 2nd equation of motion,

s = ut + 1/2at^2. (here s is the height of the building; a = g).

s = 0(3.2) + 1/2 x (10) x (3.2)^2 (since ball is dropped, initial velocity, u = 0).

s = 0 + 1/2 x 10 x 10.24 (since (3.2)^2 = 10.24)

s =  5 x 10.24

s = 51.2 m.

Ans: Therefore, the height of the building is 51.2 m.

Hope it helps!! :)