1

For any position A, show that $\tan\alpha + \tan\beta =\tan\theta$enter image description here

MY SOLUTION:

We know that $\tan\theta=\frac{4h}{R}$

$$\tan\alpha=\frac{h}{x}$$ And $$\tan\beta = \frac{h}{R-x}$$ So $$\tan\alpha + \tan\beta = \frac{hR}{xR-x^2}$$ That’s as far as I could go up to. What should I do next?

Andrei
  • 37,370
Aditya
  • 6,191
  • You are confusing $h$ in the first formula (for $\theta$) where it has the meaning of maximum height, and in later formulas, where it has the meaning of the height at position $x$ – Andrei Aug 26 '19 at 15:40
  • You are right, but I still don’t know how to proceed – Aditya Aug 26 '19 at 15:44

2 Answers2

2

You may continue by substituting the following results for a projectile into your expression

$$ x = v_0t\cos \theta $$ $$ R = \frac{v_0^2}{g}\sin2\theta $$ $$ h = v_0t\sin\theta - \frac{1}{2} g t^2 $$

and verify that

$$\frac{hR}{xR-x^2}= \tan\theta$$

The numerator is

$$hR = \left(x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}\right)\frac{v_0^2\sin 2\theta}{g}=x\tan\theta \left(\frac{v_0^2\sin 2\theta}{g}-x \right)$$

and the denominator

$$x(R-x) = x\left(\frac{v_0^2\sin 2\theta}{g}-x \right)$$

Their ratio comes out as

$$\frac{hR}{xR-x^2}= \tan\alpha + \tan\beta= \tan\theta$$

Quanto
  • 97,352
  • That doesn’t work. As Andrei, correctly pointed out, I can’t keep using h for all values, since it’s not maximum at x. So my working is wrong. I tried to do it again, but I am still stuck. – Aditya Aug 26 '19 at 15:55
  • Please see the added lines in the answer – Quanto Aug 26 '19 at 16:19
0

Let's start with the equations of motions along the two axis. In the horizontal axis you have $$x=ut\cos\theta$$ In the vertical direction you have $$h=ut\sin\theta-\frac12 gt^2$$ Taking $t=\frac{x}{u\cos\theta}$ from the first equation and plugging it into the second you get $$h=x\tan\theta-\frac12 g\frac{x^2}{u^2\cos^2\theta}$$ First, to calculate $R$ you set $h=0$. One solution is obviously $x=0$. For the other solution $$0=\tan\theta-\frac12 g\frac{R}{u^2\cos^2\theta}$$ This yields $$R=2\frac{u^2}{g}\tan\theta\cos^2\theta$$ Now rewrite the formula for the height in terms of $R$ as $$h=x\tan\theta-\frac{x^2}R\tan\theta$$ From here it should be easy, just plug this into your $\tan\alpha+\tan\beta$ equation: $$\tan\alpha+\tan\beta=\frac{hR}{x(R-x)}=\frac{xR\tan\theta-x^2\tan\theta}{x(R-x)}=\tan\theta$$

Andrei
  • 37,370
  • Didn’t understand the part where you set h=0. Why did you substitute R for x^2 – Aditya Aug 26 '19 at 16:16
  • I substituted $R$ for $x$. At $x=R$ you get $h=0$. That's the definition of the range. Note that you have an $x$ in the left term and an $x^2$ in the right term. If $x\ne 0$, you can divide by $x$ – Andrei Aug 26 '19 at 17:06