Question

If a particle is thrown upwards with a speed of 75 m/s find the distance travelled by particle in 8th sec

Answer 1

Given

A particle is thrown upwards with a speed of 75 m/s

To find

Distance travelled by the particle in the 8th second..

Explanation:

$\sf initial \: velocity \:(u)=75ms^-1$

$\sf as \:it \:is\: thrown \:vertically\: upwards \\ \sf acceleration (a)=-g$

We know that

$\sf distance\: travelled\: in\: the\: nth\: second$

$\underline{\sf S_n=u+a(n-\dfrac{1}{2})}$

Now we want the distance travelled in 8th second

$=>S_8=75+(-g)(8-\dfrac{1}{2})$

$=>S_8=75-9.8(\dfrac{16-1}{2})$

$=>S_8=75-9.8(\dfrac{15}{2})$

$=>S_8=75-4.9(15)$

$=>S_8=75-73.5=1.5m$

$=>\boxed{S_8=1.5m}$

Distance travelled by the particle in 8th second is 1.5m

Some More important formulas:

$= > v = u + at$

$= > {v}^{2} - {u}^{2} = 2as$

$=>s = ut + \dfrac{1}{2} a {t}^{2}$

Answer 2

Answer:

Explanation:

Given :-

Initial Velocity of particle , u = 75 m/s

Acceleration by the particle, a = - 9.8 m/s ( As it going upward)

To Find :-

Distance traveled in 8th second, S = ??

Formula to be used :-

S(n) = u + a (n - 1/2)

Solution :-

Putting all the values, we get

⇒ S(n) = u + a (n - 1/2)

⇒ S(8) = 75 + a (8 - 1/2)

⇒ S(8) = 75 + (- 9.8) (8 - 1/2)

⇒ S(8) = 75 - 9.8 ( 16 - 1/2)

⇒ S(8) = 75 - 9.8 (15/2)

⇒ S(8) = 75 - 4.9 (15)

⇒ S(8) = 75 - 73.2

⇒ S(8) = 1.5 m

Hence, the distance travelled by particle in 8th sec is 1.5 m.