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

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

[ Ans: Vo/2 ]

Sol:
Time of Ascent t = Vo/g

Average velocity

\displaystyle = \frac{\int_{0}^{t}V dt}{\int_{0}^{t}dt}

\displaystyle = \frac{\int_{0}^{t}(V_0 - g t)dt}{\int_{0}^{t}dt}

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

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: \displaystyle v^2 = u^2 + 2 a S

\displaystyle v^2 = 0 + 2\times 10 \times 20

v = 20 m/s

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….

Q: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. 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 – t3

Differentiating w.r.t. time

\displaystyle \frac{dx}{dt} = 0 + 12-3t^2

\displaystyle v =  12-3t^2

When v = 0 ; t = 2 sec

Again differentiating with time

\displaystyle \frac{dv}{dt} = 0-6t

\displaystyle a = -6t

a = -12 m/s2

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….

Q: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. The relation between h1, h2 and h3 is

(a) h1 = 2h2 = 3h3

(b) h1 = h2/3 = h3/5

(c) h2 = 3h1 and h3 = 3h2

(d) h1 = h2 = h3

Ans: (b)

By Applying

\displaystyle S = u t + \frac{1}{2}g t^2

\displaystyle h_1 = 0 + \frac{1}{2}g 5^2 …(i)

\displaystyle h_1 + h_2 = 0 + \frac{1}{2}g (5+5)^2 …(ii)

\displaystyle h_1 + h_2 + h_3 = 0 + \frac{1}{2}g (5+5+5)^2 …(iii)

On dividing (ii) by (i)

\displaystyle \frac{h_1 + h_2}{h_1} = 4

\displaystyle (1+\frac{h_2}{h_1}) = 4

\displaystyle \frac{h_2}{h_1} = 3

\displaystyle h_1 = \frac{h_2}{3}