Question

From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 m/s The ratio of the distance travelled by it in the 3rd and 2nd seconds of motion is ? A) 5:7 B) 7:5 C) 3:6 D)6:3

Answer 1

Answer:

(B).The ratio of the distance traveled by it in the 3rd and 2nd seconds of motion is 7:5.

Explanation:

Given that,

Velocity $v= 10\ m/s$

Using the formula for distance covered in nth second

$S_{n}=u+\dfrac{a}{2}(2n-1)$

Here, u = velocity

n= 3 rd

$a= g= 9.8 m/s^2$

$S_{3}=10 +\dfrac{9.8}{2}\times(2\times3-1)$

$S_{3}=34.5\ m$

For, n= 2nd

$S_{2}=10 +\dfrac{9.8}{2}\times(2\times2-1)$

$S_{2}=24.7\ m$

The ratio of the distance traveled by it in the 3rd and 2nd seconds of motion is

$\dfrac{S_{3}}{S_{2}}=\dfrac{34.5}{24.7}=\dfrac{35}{25}$

$\dfrac{S_{3}}{S_{2}}=\dfrac{7}{5}$

Hence, The ratio of the distance traveled by it in the 3rd and 2nd seconds of motion is 7:5.

Answer 2

Explanation:

(B).The ratio of the distance traveled by it in the 3rd and 2nd seconds of motion is 7:5.

Explanation:

Given that,

Velocity v= 10\ m/sv=10 m/s

Using the formula for distance covered in nth second

S_{n}=u+\dfrac{a}{2}(2n-1)S

n

=u+

2

a

(2n−1)

Here, u = velocity

n= 3 rd

a= g= 9.8 m/s^2a=g=9.8m/s

2

S_{3}=10 +\dfrac{9.8}{2}\times(2\times3-1)S

3

=10+

2

9.8

×(2×3−1)

S_{3}=34.5\ mS

3

=34.5 m

For, n= 2nd

S_{2}=10 +\dfrac{9.8}{2}\times(2\times2-1)S

2

=10+

2

9.8

×(2×2−1)

S_{2}=24.7\ mS

2

=24.7 m

The ratio of the distance traveled by it in the 3rd and 2nd seconds of motion is

\dfrac{S_{3}}{S_{2}}=\dfrac{34.5}{24.7}=\dfrac{35}{25}

S

2

S

3

=

24.7

34.5

=

25

35

\dfrac{S_{3}}{S_{2}}=\dfrac{7}{5}

S

2

S

3

=

5

7

Hence, The ratio of the distance traveled by it in the 3rd and 2nd seconds of motion is 7:5.