jiteshmeghwal9
  • jiteshmeghwal9
Path of a projectile projected from the ground is given as \[y=ax-bx^2\]. Find angle of projection, maximum height attained and horizontal range of the projectile.
Physics
chestercat
  • chestercat
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jiteshmeghwal9
  • jiteshmeghwal9
jiteshmeghwal9
  • jiteshmeghwal9
anonymous
  • anonymous
You can sole this with formulas your teacher provided, but I prefer solve it mathematically

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jiteshmeghwal9
  • jiteshmeghwal9
The problem is that the question has given me only one equation & has asked me to find three things.
anonymous
  • anonymous
well note that the trajectory of projectile is a parabola, can you plot that parabola for me?
jiteshmeghwal9
  • jiteshmeghwal9
|dw:1435902087173:dw|
jiteshmeghwal9
  • jiteshmeghwal9
where I have written [\a\]
anonymous
  • anonymous
you have \[y=ax-bx^2\]sry I mean \(b\)
jiteshmeghwal9
  • jiteshmeghwal9
\[b\] must be a +ve number because it is a parabolic equation or we can say quadratic
anonymous
  • anonymous
well \(b>0\) because parabola must be downward, right? It's a ball moving on the ground \(x-axis)
anonymous
  • anonymous
Can this be a path for projectile?|dw:1435902758838:dw|
jiteshmeghwal9
  • jiteshmeghwal9
How can the body move against gravity?
anonymous
  • anonymous
ok, so \(b\) can not be less than zero, because if \(b<0\) we will have a upward parabola
jiteshmeghwal9
  • jiteshmeghwal9
yes
anonymous
  • anonymous
Now, parabola (ball) hits the x-axis (ground) at \(y=0\)\[y=ax-bx^2=0\]\[x=0 \]\[x=\frac{a}{b}\]
anonymous
  • anonymous
|dw:1435903099764:dw|
jiteshmeghwal9
  • jiteshmeghwal9
|dw:1435903307350:dw|
anonymous
  • anonymous
xm is where maximum height (ym) occurs
jiteshmeghwal9
  • jiteshmeghwal9
ok
anonymous
  • anonymous
Now if you think about it mathematically and let \(f(x)=ax-bx^2\): for angle of projection: \(\tan \theta= f'(0)\) maximum height attained: \(y_m\) and horizontal range of the projectile: \(x=\frac{a}{b}\)
anonymous
  • anonymous
\[y_m=f(x_m)\]
jiteshmeghwal9
  • jiteshmeghwal9
what is \[f'(0)\] ?
anonymous
  • anonymous
derivative
anonymous
  • anonymous
have you studied derivative?
jiteshmeghwal9
  • jiteshmeghwal9
nope not yet
anonymous
  • anonymous
ok, so how you find it? do you have formula or something?
jiteshmeghwal9
  • jiteshmeghwal9
I have eqn of trajectory\[y=xtan \theta [1-\frac{ x }{ R }]\] 'R' is range.
anonymous
  • anonymous
thanks, equate \(f(x)\) and formula and find \(R\) and \(\theta\)
anonymous
  • anonymous
\[y=\tan \theta \ x - \frac{\tan \theta}{R} x^2=ax-bx^2 \]\[\tan \theta =a\]\[\frac{\tan \theta}{R}=b\]
jiteshmeghwal9
  • jiteshmeghwal9
so\[\frac{ \tan \theta }{ \frac{ \tan \theta }{ R }}=\frac{ a }{ b }\]\[R=\frac{ a }{ b }\]also\[\tan \theta=a\]\[\theta =\tan^{-1} a\]
jiteshmeghwal9
  • jiteshmeghwal9
got it, thanks friend :)

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