A particle is thrown vertically upwards then its velocity so that it . . .

Question : A particle is thrown vertically upwards then its velocity so that it covers same distance in 5th and 6th second would be :

(a) 48 m/s

(b) 14 m/s

(d) 7 m/s

Doubt by Girisha

Solution :

Before doing this question, we have to understand that the vertical motion under gravity is an accelated motion so it is not possible that the distance covered by the body is consecutive seconds should be equal to each other, except the case when direction of motion is opposite in both the case.

So we have consider that the body is through up and it reaches the highest point at time = 5 sec.

Only then the distance covered in the 5th second (upward motion) would be equal to the distance covered in the 6th second (downward motion).

This question futher can be done by two methods either by using foruth equation of motion or without it.

Solution by using the fourth equation of motion

We know,
Snth = u+[2n-1]g/2
S5th=u+[2(5)-1](-g/2)
S5th=u-9g/2 — (1)

Also
S6th=u+[2(6)-1](-g/2)
S6th=u-11g/2 — (2)

Now,
|S5th|=|S6th|
But both the displacements are in the opposite direction so
S5th=-S6th — (3)

Using (1), (2) and (3)

-(u-9g/2)=u-11g/2
-u+9g/2=u-11g/2
9g/2+11g/2=u+u
20g/2=2u
10g=2u
5g=u
u=5g
u=5×9.8
u=49 m/s

Solution by not using the fourth equation of motion

We know,

The body reaches the highest point at t=5 sec

At the highest point the velocity is zero (v=0)

Let initial velocity be u.

Using
v=u+at

0=u+(-g)(5)
-u=-5g
u=5g

u=5×9.8
u=49 m/s

Hence, (c) 49 m/s, would be the correct option.