Science of Apps! - Part 4: Fruit Ninja

To solve x

the first object, watermelon, goes up with velocity x, reaches a max height of 5m, and then comes down. So,at a distance of 1m from the ground, its velocity while going up will be equal to its velocity while coming down. ( This is bcoz we are considering an ideal case and there is no other thing like wind flow, air resistance etc. that can affect its velocity ).

Now, let velocity( while going up & while coming down bcoz they are equal ) at 1m distance from the ground = v , then considering the case of going up

$v^{2}$ - $x^{2}$ = - 2gh = $-\quad 2*10*1$ = -20 . The negative sign on the right hand side is there bcoz the object is going upwards while the acceleration is acting downwards

now considering the case of watermelon coming down, we know that the its velocity is zero when it is at its maximum height, therefore this time taking the initial velocity to be zero, we get

$v^{2}$ - $0^{2}$ = $2gh\quad =\quad 2*10*4\quad =\quad 80$ . The positive sign on the right hand side is there bcoz the object is going downwards and the acceleration is also acting downwards

Placing this value in the previous equation, we get

$x^{2}$ = 80 + 20 = 100 => x=10

To solve y

Now, when the watermelon is at its peak, a coconut is thrown upwards at the same point of time. So, now our watermelon is coming down and the coconut is going up. and they are on the same horizontal line at 1m height from the ground.

SO, to reach the 1m height from the ground at the same time

time taken by watermelon to reach this height from its maximum must be equal to the time taken by coconut to reach this height from ground level.

now, calculating the time taken by watermelon

$s\quad =\quad ut\quad +\quad (1/2)a (t^{2})$

$4\quad =\quad 0\quad +(1/2)*10(t^{ 2})$

$\Longrightarrow \quad t^ { 2} \quad =\quad 4/5$

$\Longrightarrow \quad t\quad =\sqrt { \frac { 4 }{ 5 } } \quad =\quad \frac { 2 }{ \sqrt { 5 } }$

now, calculating the time taken by coconut to go up to a height of 1m from ground level

$1\quad =\quad yt\quad -\quad (1/2)*10*(t^{2})$

Placing the value of t from the previous equation, we get

$y\quad =\quad \frac { 5\sqrt { 5 } }{ 2 } \quad =\quad 5.59$

Now, we have the values of x and y. But the question asks us to find the greatest integer values of x and y.

So, greatest integer value of x = [x] = [10] = 10 and greatest integer value of y = [y] = [ 5.59 ] = 5

and 10/5 =2 and so 2 is our right answer.

The time $t$ for the watermelon to reach the apex at 5 m is given by: $t=\frac{x}{g} = 0.1x$ And it is also known that: $5=xt-\frac{1}{2}gt^2=0.1x^2 - 5(0.01)x^2=0.05x^2$ $\Rightarrow x=\sqrt{\frac{5}{0.05}} = 10 m/s$ The time $t_1$ taken for the watermelon to drop to 4 m to meet the coconut at 1 m above ground is given by: $4=\frac{1}{2}gt_1^2 \Rightarrow t_1=\sqrt{\frac{4}{5}}$ It is also the time for the coconut to reach 1 m above ground, therefore, $1=yt_1-\frac{1}{2}gt_1^2$ $\Rightarrow y=\frac{(1+4)}{\sqrt{\frac{4}{5}}}= \frac{5\sqrt{5}}{2}=5.59017$ Therefore, $\frac { \left\lfloor x \right\rfloor }{ \left\lfloor y \right\rfloor } =\frac{10}{5}= \boxed{2}$

from the top most portion the watermelon drops freely to a height of 1m initial velocity is zero so we get time t. the coconut is going up with initial velocity y, displacement 1m from this we get y. now applying conservation of energy to melon, x can also be found out.

Well, the solution is actually 1.7888 ... But as it only takes integer as answer so I rounded it up to 2...