Motion in A Straight Line Test - 18

Given,

Object is free fall from height $$h$$

Initial velocity, $$u=0$$

Apply kinematic equation of motion

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

$$ h=\dfrac{1}{2}g{{t}^{2}}\,\,........\,(1) $$

At half time

$$s=u\dfrac{t}{2}+\dfrac{1}{2}g{{\left( \dfrac{t}{2} \right)}^{2}}=\dfrac{1}{4}\left( \dfrac{1}{2}g{{t}^{2}} \right)=\dfrac{h}{4}$$

Height from ground, $$h-\dfrac{h}{4}=\dfrac{3h}{4}$$meter

Hence, at half time height from ground is $$\dfrac{3h}{4}$$m

suppose $$n - balls \to $$

$$final\,velocity\,,\,$$ $$v = 0$$

$$u = 0$$

$$T = {\dfrac{1} {n}}\sec $$

$$0 = u - g{\dfrac{1}{n}}$$

$${\dfrac{g} {n}} = u$$

$${v^2} = {u^2} - 2gh$$

$$0 = {\dfrac{{g^2}} {{n^2}}} - 2gh$$

$${\dfrac{{g^2}} {{n^2}}} = 2gh$$

$$h = {\dfrac{g} {2{n^2}}}$$

$$n = 5$$

   $$ = {\dfrac{10} {2 \times 25}}$$

   $$ = {\dfrac{1}{5}}m$$

Given that,

Height $$=h$$

Time $$t=\dfrac{3T}{4}$$

Since, the ball is released from rest, its initial velocity is zero. When released from height h it reaches the ground in time T.

If we apply the second equation of motion

  $$ s=ut+\dfrac{1}{2}g{{t}^{2}} $$

 $$ h=\dfrac{g{{T}^{2}}}{2} $$

 $$ {{T}^{2}}=\dfrac{2h}{g} $$

Now, we apply the same equation to calculate the height covered in time 3T/4

  $$ s=ut+\dfrac{1}{2}g{{t}^{2}} $$

 $$ s=g\times \dfrac{{{\left( \dfrac{3T}{4} \right)}^{2}}}{2} $$

 $$ s=\left( \dfrac{9g}{32} \right)\times {{T}^{2}} $$

 $$ s=\left( \dfrac{9g}{32} \right)\times \dfrac{2h}{g} $$

 $$ s=\dfrac{9h}{16} $$

Now, the distance from the ground

  $$ s=h-\dfrac{9h}{16} $$

 $$ s=\dfrac{7h}{16} $$

Hence, the distance from the ground is $$\dfrac{7h}{16}$$ 

The equation of motion is

$$\vec{v}=\vec{u}+\vec{a}t$$

$$or\,\,\vec{v}=-\vec{u}-\vec{a}t$$

Direction of initial velocity and direction of acceleration should be same, either negative or positive

Hence, at negative velocity and negative acceleration object will speed up

Given,

Initial velocity, $$u=0$$

Time period of free fall from height $$125\,m$$

$$t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 125}{10}}=5\,\sec $$

Apply kinematic equation of motion

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

Ratio of displacement for $${{1}^{st}}$$ sec and last $$1\,\sec $$

$$\dfrac{{{s}_{1}}-{{s}_{0}}}{{{s}_{4}}-{{s}_{5}}}=\dfrac{\dfrac{1}{2}g\left( {{1}^{2}}+0 \right)}{\dfrac{1}{2}g\left( {{5}^{2}}+{{4}^{2}} \right)}=\dfrac{1}{9}$$

Hence , ratio is $$1:9$$  

 $$v=ut+\dfrac{1}{2}\times gt^2$$

Differentiate y with respect to t which gives velocity as dy/dt = v

$$\dfrac{dv}{dt}$$ = $$u + \dfrac{1}{2}g (2t)$$

v=u + gt -------------------Eqn (1) 

Differentiate v with respect to t which gives velocity as d2y/dt2 = dv/dt = a

$$\dfrac{dv}{dt}$$ = 0 + g

a = g  ----------------------Eqn (2)We know that force acting on a mass m is given by F = ma

Now from Eqn (2), substitute a =g, Hence, F = mg

C. Acceleration

Given,

Initial velocity, $$u=5\,m{{s}^{-1}}$$

Acceleration, $$a=2\,m{{s}^{-2}}$$

Apply kinematic equation of motion

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

$$ s=5\times 10+\dfrac{1}{2}\times 2\times 10^2=150\,m $$

Hence in first 10 second card travel $$150\,m$$

Statement D is correct
Situation 1:
We want time to cross x.
$$v^{2}=u^{2}+2as$$
$$u=0, a=-g, s=-1 $$
$$v^{2}=2g\Rightarrow v_{1}=\sqrt{2g}$$
$$(1+x)=\dfrac{1}{2} gt^{2}\Rightarrow

T_{2}=\sqrt{\dfrac{2(1+x)}{g}}$$
$$1=\dfrac{1}{2}gt^{2}=T_{1}=\sqrt{\dfrac{2}{g}}$$
So time taken to cross $$x=t_{1}=T_{2}-T_{1}

=\sqrt{\dfrac{2}{g}}(\sqrt{1+x}-1)$$
Situation 2:
$$(2+x)=\dfrac{1}{2}gt^{2}\Rightarrow

T_{2}=\sqrt{\dfrac{2(2+x)}{g}}$$
$$2=\dfrac{1}{2}gt^{2}\Rightarrow T_{1}=\sqrt{\dfrac{2\times 2}{g}}$$
So time taken to cross $$x=t_{2}=T_{2}-T_{1}=\sqrt{\dfrac{2}{g}}(\sqrt{2+x}-\sqrt{2})

$$
Clearly $$t_{2}< t_{1} $$