Question

A particle is thrown vertically upward with a velocity 48
m/s. If g = 10 m/s, then the distance travelled by
particle during 5th second is
1.5 m
3 m
1.7 m
3.4 m​

Answer 1

velocity 48 ,g=10 m/s ,sec is 3.4

Answer 2

The distance travelled by  particle during 5th second is 7m

Explanation:

Given:

Velocity of the particle in upward direction = 48 m/s

To find out:

The distance travelled during the 5th second

Solution:

We'll use the second equation of motion

$s=ut+\frac{1}{2}gt^2$

Initial velocity

$u=-48$ m/s

The distance travelled during 5th second

= Distance travelled in 6 seconds - Distance travelled in 5 seconds

$=-48\times 6+\frac{1}{2}\times 10\times 6^2-(-48\times 5+\frac{1}{2}\times 10\times 5^2)$

$=48(-6+5)+5(6^2-5^2$

$=-48+5\times (36-25)$

$=-48+5\times 11$

$=-48+55$

$=7$ m

Hope this answer is helpful.

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