Motion in A Straight Line Test - 52

Given,

Initial velocity $$=u$$

After $$2\ cm\,$$penetration, velocity $$v=\dfrac{u}{2}$$

Apply equation of kinematic

 $$ {{v}^{2}}-{{u}^{2}}=2as $$

 $$ {{\left( \dfrac{u}{2} \right)}^{2}}-{{u}^{2}}=2a\times 2 $$

 $$ a=\dfrac{-3{{u}^{2}}}{16} $$

When, Final velocity $${{v}_{1}}=\dfrac{u}{4}$$

Apply kinematic equation of motion

 $$ v_{1}^{2}-{{u}^{2}}=2a{{s}_{1}} $$

 $$ {{\left( \dfrac{u}{4} \right)}^{2}}-{{u}^{2}}=2\left( \dfrac{-3{{u}^{2}}}{16} \right){{s}_{1}} $$

 $$ {{s}_{1}}=2.5\ cm $$

Hence, for final velocity to be one fourth of initial velocity, body penetrate $$0.5\ cm$$ more in wood.

Given that,

Initial velocity $$u=20\,m/s$$

Final velocity $$v=0$$

Now, from equation of motion

  $$ v=u-gt $$

 $$ 0=20-10t $$

 $$ {{t}_{1}}=\dfrac{20}{10} $$

 $$ {{t}_{1}}=2\,s $$

Total time

 $$ t=2{{t}_{1}} $$

 $$ t=2\times 2 $$

 $$ t=4\,s $$

Hence, the time is $$4\ s$$

Initial velocity $$u=20\,m/s$$

Final velocity $$v=0\,m/s$$

Acceleration $$a=-10\,m/{{s}^{2}}$$

Now, the time by ball reach its maximum height

From equation of motion

  $$ v=u+at $$

 $$ 0=20-10t $$

 $$ t=2\,s $$

So, the time taken by the same ball to return to the hands of the juggler is  

  $$ =\dfrac{2u}{g} $$

 $$ =\dfrac{2\times 20}{10} $$

 $$ =4\,s $$

So, he is throwing the balls after 1 sec each,

Let at some instant he throws ball number 4.

 Now, Before 1 s of throwing it, he throws ball 3.

So, the height of ball 3 is

  $$ {{h}_{3}}=20\times 1-\frac{1}{2}\times 10\times {{\left( 1 \right)}^{2}} $$

 $$ {{h}_{3}}=15\,m $$

Before 2s, he throws ball 2, so the height of ball 2 is

  $$ {{h}_{2}}=20\times 2-\frac{1}{2}\times 10\times 4 $$

 $$ {{h}_{2}}=20\,m $$

Before 3s, he throws ball 1, so the height of ball 1 is

  $$ {{h}_{1}}=20\times 3-\frac{1}{2}\times 10\times 3\times 3 $$

 $$ {{h}_{3}}=15\,m $$

Hence, the height of balls in air from the ground will be $$15\ m$$, $$20\ m$$, $$15\ m$$ and $$0\ m$$