A body is projected vertically up with a speed V_{o} . Find the magnitude of time average velocity of the body during its ascent.

[ Ans: V_{o}/2 ]

Sol:

Time of Ascent t = V_{o}/g

Average velocity

Skip to content
# Tag: Kinematics

## A body is projected vertically up with a speed Vo . Find the magnitude of time average velocity of the body during its ascent.

## A boy standing, at the top of a tower of 20 m height drops a stone. Assuming, g = 10 ms–2, the velocity with which it hits the ground is

## The motion of a particle along a straight line is described by equation x = 8 + 12t – t3 where, x is in metre and t in sec….

## A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5s, the next 5s and the next 5s respectively….

A body is projected vertically up with a speed V_{o} . Find the magnitude of time average velocity of the body during its ascent.

[ Ans: V_{o}/2 ]

Sol:

Time of Ascent t = V_{o}/g

Average velocity

Question: A boy standing, at the top of a tower of 20 m height drops a stone. Assuming, g = 10 ms^{–2}, the velocity with which it hits the ground is

(a) 20 m/s

(b) 40 m/s

(c) 5 m/s

(d) 10 m/s

**Ans: (a)**

Sol:

v = 20 m/s

Q:The motion of a particle along a straight line is described by equation

x = 8 + 12t – t^{3 } . where, x is in metre and t in sec. The retardation of the particle when its velocity becomes zero is

(a) 24 ms^{–2}

(b) zero

(c) 6 ms^{–2}

(d) 12 ms^{–2}

**Ans: (d)**

Sol: x = 8 + 12t – t^{3 }

Differentiating w.r.t. time

When v = 0 ; t = 2 sec

Again differentiating with time

a = -12 m/s^{2}

Q:A stone falls freely under gravity. It covers distances *h*_{1}, *h*_{2} and *h*_{3} in the first 5s, the next 5s and the next 5s respectively. The relation between *h*_{1}, *h*_{2} and *h*_{3} is

(a) h_{1} = 2h_{2} = 3h_{3}

(b) h_{1} = h_{2}/3 = h_{3}/5

(c) h_{2} = 3h_{1} and h_{3} = 3h_{2}

(d) h_{1} = h_{2} = h_{3}

**Ans: (b)**

By Applying

…(i)

…(ii)

…(iii)

On dividing (ii) by (i)