Cody gives a chocolate ...

Classical Mechanics Level 3

Cody is going out , and his GF is angry for this .... So Cody tries to convince her , but for that he doesn't have much time to go to her . So he throws a chocolate towards her ... (thinking that she might forgive him :P ) . He threw it with initial velocity $u=49 m/s$ and making an angle $\theta = 45 ^{\circ}$ with the horizontal and it was a perfect throw so it was going to reach the target (his GF) . But suddenly , an unexpected blow of wind came , caused an acceleration of $a=5 m/s^2$ in horizontal direction. So the chocolate went somewhat far away from his GF , so his GF had to run a distance to get the chocolate . How much distance in $meters$ did she have to run to get the chocolate ?

Details and Assumptions :-

Actual expected path of chocolate is shown with black dotted path and the path which was caused due to wind is shown as red line. (The red dot means chocolate)
acceleration due to gravity= $g=9.8 m/s^2$

The answer is 125.

1 solution

Aditya Raut
Jun 2, 2014

As we can see , the original distance expected by the chocolate to travel horizontally is

$S_x = u_x\times t$ as $a$ was expected as 0.

the vertical distance is $S_y=0=u_yt - \frac{1}{2} g t^2$ ..and t is not zero (t is time)

Hence we get that $\frac{2u_y}{g}=t$ and then , because angle is 45 degree, we get $u_x=u_y= \frac{49}{\sqrt{2}}$ thus $t= 5\sqrt{2} second$ .

$S_x= u_x \times t = 245 m$ but because of the wind's $a=5m/s^2$ , this distance will increase.

$S_x = 245 + \frac{1}{2} a t^2 = 245 + \frac{5\times 50}{2} = 245 +125$

Thus Cody's GF will have to run $\boxed{125}$ meters to collect the chocolate !

Cody's GF has to run at the speed of a car.. :P

Satvik Golechha - 6 years, 9 months ago

Horizontal and vertical components of velocities are equal say V. Let .. t s be the time to reach GF. $\\V=49*sin(45^o).~ \therefore~V=\dfrac{g*t}{2} \implies t=\dfrac{10}{\sqrt 2}$ The required difference is only due to acceleration of $~5~ m/s^2$ . Thus the distance to run back is D given by $D=\dfrac{1}{2}*acc*t^2=\dfrac{1}{2}*5*(\dfrac{10}{\sqrt 2})^2=125 m.$

Niranjan Khanderia - 6 years, 9 months ago

I dont think that his GF will run such a long distance for a single choco..!! =D

Deepanshu Gupta - 6 years, 9 months ago

it would be a tiresome task for cody's gf to catch the choclate

Piyush Bharti - 6 years, 9 months ago

Well, computing the 245 part is redundant

Agnishom Chattopadhyay - 6 years, 8 months ago

Cody gives a chocolate ...

The answer is 125.

1 solution

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