Question

Ahola♥️♥️



A particle starts droped from some height in last second of motion it travels a distance which is equal to distance travelled initial 5 second then find out time taken by particle to strike at ground.

Answer 1

The Given Information is

t in the last second = 1 s

s in the last second = x m

u = 0 m/s. (Since the stone falls from the cliff with no velocity)

a=g = 9.8 m/s²

Now,

v = s/t

v = x m/1s
v = x m/s

But this is the average velocity in the last second.

With the the acceleration of 9.8 m/s², we can calculate the final velocity.

V²-U²=2as

(V+U)(V-U)=2as

We know, v=u+at

Therefore v-u =at

Substituting in previous equation.

(V+U)(at)=2as

(V+U)=68.6m/s (After substituting the values)
Solving for V, we get
 V=39.2m/s

Now according to Newton's laws of motion,

V²-u²= 2as

(x m/s)²-(0)²= 2(9.8)(S)

(x)/19.6= S
 


Note:
>v-u = at = 9.8 use this in solving for V

Answer 2

Hey mate here is your answer

The Given Information is

t in the last second = 1 s

s in the last second = x m

u = 0 m/s. (Since the stone falls from the cliff with no velocity)

a=g = 9.8 m/s²

Now,

v = s/t

v = x m/1s

v = x m/s

But this is the average velocity in the last second.

With the the acceleration of 9.8 m/s², we can calculate the final velocity.

V²-U²=2as

(V+U)(V-U)=2as

We know, v=u+at

Therefore v-u =at

Substituting in previous equation.

(V+U)(at)=2as

(V+U)=68.6m/s (After substituting the values)

Solving for V, we get

 V=39.2m/s

Now according to Newton's laws of motion,

V²-u²= 2as

(x m/s)²-(0)²= 2(9.8)(S)

(x)/19.6= S

 

Note:

>v-u = at = 9.8 use this in solving for V

Thanks