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jiteshmeghwal9
 one year ago
Path of a projectile projected from the ground is given as \[y=axbx^2\]. Find angle of projection, maximum height attained and horizontal range of the projectile.
jiteshmeghwal9
 one year ago
Path of a projectile projected from the ground is given as \[y=axbx^2\]. Find angle of projection, maximum height attained and horizontal range of the projectile.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can sole this with formulas your teacher provided, but I prefer solve it mathematically

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0The problem is that the question has given me only one equation & has asked me to find three things.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well note that the trajectory of projectile is a parabola, can you plot that parabola for me?

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435902087173:dw

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0where I have written [\a\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have \[y=axbx^2\]sry I mean \(b\)

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0\[b\] must be a +ve number because it is a parabolic equation or we can say quadratic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well \(b>0\) because parabola must be downward, right? It's a ball moving on the ground \(xaxis)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can this be a path for projectile?dw:1435902758838:dw

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0How can the body move against gravity?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so \(b\) can not be less than zero, because if \(b<0\) we will have a upward parabola

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, parabola (ball) hits the xaxis (ground) at \(y=0\)\[y=axbx^2=0\]\[x=0 \]\[x=\frac{a}{b}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435903099764:dw

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435903307350:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0xm is where maximum height (ym) occurs

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now if you think about it mathematically and let \(f(x)=axbx^2\): for angle of projection: \(\tan \theta= f'(0)\) maximum height attained: \(y_m\) and horizontal range of the projectile: \(x=\frac{a}{b}\)

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0what is \[f'(0)\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0have you studied derivative?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so how you find it? do you have formula or something?

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0I have eqn of trajectory\[y=xtan \theta [1\frac{ x }{ R }]\] 'R' is range.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks, equate \(f(x)\) and formula and find \(R\) and \(\theta\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[y=\tan \theta \ x  \frac{\tan \theta}{R} x^2=axbx^2 \]\[\tan \theta =a\]\[\frac{\tan \theta}{R}=b\]

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0so\[\frac{ \tan \theta }{ \frac{ \tan \theta }{ R }}=\frac{ a }{ b }\]\[R=\frac{ a }{ b }\]also\[\tan \theta=a\]\[\theta =\tan^{1} a\]

jiteshmeghwal9
 one year ago
Best ResponseYou've already chosen the best response.0got it, thanks friend :)
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