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An object which is projected from the ground with an angle of $\theta$ passes two walls(without touching) within its motion.The wall which has a height of 'a' is situated in a distance of 'v' from the projected place and the other wall which has a height of 'b' is situated in a distance of 'a' from the first wall. Find a expression with a and b for the $\tan$ value of the $\theta$.

Parabolic Trajectory

I tried to find the $\theta$,

$\rightarrow s = ut \\ v=u\cos\theta.t_{1}\\$

$\uparrow s=ut+\frac{1}{2}at^{2}\\ a=u\sin\theta\times\frac{v}{ucos\theta}-\frac{g}{2}\times\frac{v^{2}}{u^{2}\cos^{2}\theta}\\ a=v \tan\theta-\frac{gv^{2}\sec^{2}}{2u^{2}}$

$\rightarrow s=ut\

v+a = u\cos\theta\times t_{2}$

$\uparrow s=ut+\frac{1}{2}at^{2}\\ b=u\sin\theta\times\frac{v+a}{u\cos\theta}-\frac{g}{2}\times\frac{(v+a)^{2}}{u^{2}\cos^{2}\theta}\\ b=(v+a)\tan\theta-\frac{g(v+a)^{2}\sec^{2}\theta}{2u^{2}}$

I don't see a way to find the $\theta$.Please help.

Narasimham
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Nimantha
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1 Answers1

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Do not complicate things. Just handle a parabola through the origin .

$$ y= A x -B x^2 $$ $ u= v+a $

Point $(v,a)$ $$ a = Av - Bv^2 \tag{1}$$ Point $ (u,b)$ $$ b = Au- Bu^2 \tag{2}$$

Solve (1),(2) for $ A,B $. You can take it further.

Slope is obtained by differentiating $y(x)$ and finding derivative at (0,0)

$$ dy/dx = A . $$

Narasimham
  • 40,495