Question

A particle dropped from a tower covers 5m in1 second. How much distance will it cover in 2second?​

Answer 1

Answer:

15

Explanation:

distance covered in nth second = u + 1/2(a(2n-1))

now put a=g n=1 you will get u=0

now again put u=0 a=g n= 2

you will get distance travelled in 2nd second = 15

Answer 2

Answer:-

h = 15m

Given :-

h = 5 m

t = 1s

g = -10 m/s²

t = 2s

To find :-

The distance covered in 2 second.

Solution:-

$\text{when a body is dropped from a height it's initial velocity becomes i = 0 }$

$\text{Let it cover h distance}$

By using 2nd equation of motion:-

$\huge \boxed {h = ut - \dfrac{1}{2}gt^2}$

Put the given value,

→$h = 0 - \dfrac{1}{2}\times -10 \times 2^2$

→$h = -(-5) \times 4$

→$h = 5 \times 4$

→$h = 20 m$

Distance cover in 2s will be:-

= 20 - 5

20 - 5= 15 m