Anti-Gravity in projectiles

Classical Mechanics Level 5

In the vertical xy plane a particle is projected with a velocity 10 m / s 10 m/s making an angle 45 0 { 45 }^{ 0 } with the + v e +ve x-axis. Acceleration due to gravity is − 10 j m / s 2 -10j m/{ s }^{ 2 } .

At time t = α t=\alpha sec the anti-gravity switch is turned on. Now the acceleration is 10 j m / s 2 10j m/{ s }^{ 2 } . If the particle passes through the point ( 30 , 20 ) (30, 20) during it's motion then: Find α \alpha .


The answer is 1.414.

Ronak Agarwal by Ronak Agarwal , 23, India

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3 solutions

Ronak Agarwal
Jun 9, 2014

u x = 5 2 m / s { u }_{ x }=\quad 5\sqrt { 2 } m/s

Time taken for the particle to reach x = 30 m x=30\quad m is 30 5 2 = 3 2 s e c \frac { 30 }{ 5\sqrt { 2 } } =\quad 3\sqrt { 2 }\quad sec

Now distance travelled in y-direction before the start of anti-gravity :

s y 1 = 5 2 α − 5 α 2 { s }_{ { y }_{ 1 } } = 5\sqrt { 2 } \alpha -5{ \alpha }^{ 2 }

Also at t = α , v y = 5 2 − 10 α t=\alpha,{ v }_{ y }= 5\sqrt { 2 } -10\alpha

Now distance travelled in y-direction till the start of anti gravity to t = 3 2 t=3\sqrt { 2 } sec is :

s y 2 = ( 5 2 − 10 α ) ( 3 2 − α ) + 5 ( 3 2 − α ) 2 { s }_{ { y }_{ 2 } } = (5\sqrt { 2 } -10\alpha )(3\sqrt { 2 } -\alpha )+5{ (3\sqrt { 2 } -\alpha ) }^{ 2 }

Now s y 1 + s y 2 = 20 { s }_{ { y }_{ 1 } }+{ s }_{ { y }_{ 2 } } = 20

Putting the values we get 10 α 2 − 60 2 α + 120 = 20 10{ \alpha }^{ 2 }-60\sqrt { 2 } \alpha +120=20

Solving the quadratic we get :

α = 2 o r 5 2 s e c \alpha = \sqrt { 2 } \quad or\quad 5\sqrt { 2 } \quad sec

But α < 3 2 \alpha <3\sqrt { 2 } hence α = 2 s e c . \boxed { \alpha =\sqrt { 2 } sec. }

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i got the same equations ... its just that i took the approximation ( approx. value ) and ultimately ended up getting 1.5 (this too approx. :P ) ... I thought the calculation we keep on becoming more and more complex and be time consuming but your solution is just the perfect one ...

PS:- Good thinking (question) indeed .. now i feel my mistake but proud to be conceptually correct :D

Vishruth khare - 5 years, 11 months ago

a particle is fired vertically upward with a speed of 9.8 km/s find the maxium height attained by the particle radius of earth is 6400km ang g at the surface 9.8m/s2 coinser the gravity

shivam shukla - 7 years ago

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20906 k m a p p r o x i m a t e l y . T h i s p r o b l e m c a n b e e a s i l y d o n e c o n s i d e r i n g t h e c h a n g e i n k i n e t i c e n e r g y a n d p o t e n t i a l e n e r g y g a i n e d . P o t e n t i a l e n e r g y g a i n e d = m g R e 2 ( 1 R e − 1 ( R e + h ) ) . K i n e t i c e n e r g y l o s t = 1 2 m v 2 . E q u a t i n g t h e m w e g e t 1 2 m v 2 = m g R e 2 ( 1 R e − 1 ( R e + h ) ) . N o w s o l v e f o r h . 20906\quad km\quad approximately.\quad This\quad problem\quad can\quad be\quad easily\quad done\\ considering\quad the\quad change\quad in\quad kinetic\quad energy\quad and\quad potential\\ energy\quad gained.\quad Potential\quad energy\quad gained\quad =\quad mg{ R }_{ e }^{ 2 }(\frac { 1 }{ { R }_{ e } } -\frac { 1 }{ { (R }_{ e }+h) } ).\\ Kinetic\quad energy\quad lost\quad =\quad \frac { 1 }{ 2 } m{ v }^{ 2 }.Equating\quad them\quad we\quad get\\ \frac { 1 }{ 2 } m{ v }^{ 2 }=mg{ R }_{ e }^{ 2 }(\frac { 1 }{ { R }_{ e } } -\frac { 1 }{ { (R }_{ e }+h) } ).Now\quad solve\quad for\quad h.

Ronak Agarwal - 7 years ago

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can two particles be in equilibrium under the action of their mutual gravitational force can 3 particles be can one of the 3 particle be

shivam shukla - 6 years, 12 months ago

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@Shivam Shukla – Please make your question clear

Ronak Agarwal - 6 years, 12 months ago

By the way have you tried my question

Ronak Agarwal - 7 years ago

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is radius of gyration is a constant value

shivam shukla - 6 years, 6 months ago

i did exactly as you had done only i made a sign error..

Siddharth Tiwari - 6 years, 10 months ago

Think this is the easiest problem of the set but the interesting one too....Thanks @Ronak

Saurabh Patil - 6 years, 1 month ago

i would have got the answer right if i wouldn't have read x=20 . i misread the co-ordinates

Deepansh Jindal - 4 years, 10 months ago

Solve [ 20 = − 5 α 2 + 5 ( 3 2 − α ) 2 + ( 10 2 − 10 α ) ( 3 2 − α ) + 10 α 2 ∧ 0 ≤ α ≤ 3 2 , α , R ] ⇒ 2 \text{Solve}\left[20=-5 \alpha ^2+5 \left(3 \sqrt{2}-\alpha \right)^2+\left(\frac{10}{\sqrt{2}}-10 \alpha \right) \left(3 \sqrt{2}-\alpha \right)+\frac{10 \alpha }{\sqrt{2}}\land 0\leq \alpha \leq 3 \sqrt{2},\alpha ,\mathbb{R}\right] \Rightarrow \sqrt 2 . 20 20 is the desired final height. 3 2 3\sqrt 2 the flight time. This provides the restriction of α \alpha 's value. The left portion of the right hand expression is the height gained during the initial phase and the right portion is the height gained during the final phase.

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Aryan Goyat
Sep 27, 2015

TIME OF FLIGHT IS ROOT 2 , SO WHATEVER YOU GOT MATHEMATICALLY IS WRONG AS PARTICLE GET INTO NEGATIVE 4th QUADRANT WHICH IS NOT POSSIBLE.

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https://brilliant.org/problems/want-to-visit-mit/ try

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