## jiteshmeghwal9 one year ago 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.

1. jiteshmeghwal9

@Michele_Laino

2. jiteshmeghwal9

@mukushla

3. anonymous

You can sole this with formulas your teacher provided, but I prefer solve it mathematically

4. jiteshmeghwal9

The problem is that the question has given me only one equation & has asked me to find three things.

5. anonymous

well note that the trajectory of projectile is a parabola, can you plot that parabola for me?

6. jiteshmeghwal9

|dw:1435902087173:dw|

7. jiteshmeghwal9

where I have written [\a\]

8. anonymous

you have $y=ax-bx^2$sry I mean $$b$$

9. jiteshmeghwal9

$b$ must be a +ve number because it is a parabolic equation or we can say quadratic

10. anonymous

well $$b>0$$ because parabola must be downward, right? It's a ball moving on the ground $$x-axis) 11. anonymous Can this be a path for projectile?|dw:1435902758838:dw| 12. jiteshmeghwal9 How can the body move against gravity? 13. anonymous ok, so \(b$$ can not be less than zero, because if $$b<0$$ we will have a upward parabola

14. jiteshmeghwal9

yes

15. anonymous

Now, parabola (ball) hits the x-axis (ground) at $$y=0$$$y=ax-bx^2=0$$x=0$$x=\frac{a}{b}$

16. anonymous

|dw:1435903099764:dw|

17. jiteshmeghwal9

|dw:1435903307350:dw|

18. anonymous

xm is where maximum height (ym) occurs

19. jiteshmeghwal9

ok

20. 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: $$x=\frac{a}{b}$$

21. anonymous

$y_m=f(x_m)$

22. jiteshmeghwal9

what is $f'(0)$ ?

23. anonymous

derivative

24. anonymous

have you studied derivative?

25. jiteshmeghwal9

nope not yet

26. anonymous

ok, so how you find it? do you have formula or something?

27. jiteshmeghwal9

I have eqn of trajectory$y=xtan \theta [1-\frac{ x }{ R }]$ 'R' is range.

28. anonymous

thanks, equate $$f(x)$$ and formula and find $$R$$ and $$\theta$$

29. anonymous

$y=\tan \theta \ x - \frac{\tan \theta}{R} x^2=ax-bx^2$$\tan \theta =a$$\frac{\tan \theta}{R}=b$

30. jiteshmeghwal9

so$\frac{ \tan \theta }{ \frac{ \tan \theta }{ R }}=\frac{ a }{ b }$$R=\frac{ a }{ b }$also$\tan \theta=a$$\theta =\tan^{-1} a$

31. jiteshmeghwal9

got it, thanks friend :)