Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).
At time
, the horizontal position
and vertical position
of the ball are given respectively by
![x = v_i \cos(\theta_i) t](https://tex.z-dn.net/?f=x%20%3D%20v_i%20%5Ccos%28%5Ctheta_i%29%20t)
![y = v_i \sin(\theta_i) t - \dfrac g2 t^2](https://tex.z-dn.net/?f=y%20%3D%20v_i%20%5Csin%28%5Ctheta_i%29%20t%20-%20%5Cdfrac%20g2%20t%5E2)
and the horizontal velocity
and vertical velocity
are
![v_x = v_i \cos(\theta_i)](https://tex.z-dn.net/?f=v_x%20%3D%20v_i%20%5Ccos%28%5Ctheta_i%29)
![v_y = v_i \sin(\theta_i) - gt](https://tex.z-dn.net/?f=v_y%20%3D%20v_i%20%5Csin%28%5Ctheta_i%29%20-%20gt)
The ball reaches its maximum height with
. At this point, the ball has zero vertical velocity. This happens when
![v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g](https://tex.z-dn.net/?f=v_i%20%5Csin%28%5Ctheta_i%29%20-%20gt%20%3D%200%20%5Cimplies%20t%20%3D%20%5Cdfrac%7Bv_i%20%5Csin%28%5Ctheta_i%29%7Dg)
which means
![y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g](https://tex.z-dn.net/?f=y%20%3D%20%5Cdfrac%20R6%20%3D%20v_i%20%5Csin%28%5Ctheta_i%29%20%5Ctimes%20%5Cdfrac%7Bv_i%20%5Csin%28%5Ctheta_i%29%7Dg%20-%20%5Cdfrac%20g2%20%5Cleft%28%5Cdfrac%7Bv_i%20%5Csin%28%5Ctheta_i%29%7Dg%5Cright%29%5E2%20%5C%5C%5C%5C%20%5Cimplies%20R%20%3D%20%5Cdfrac%7B6%7Bv_i%7D%5E2%20%5Csin%5E2%28%5Ctheta_i%29%7Dg%20-%20%5Cdfrac%7B3%7Bv_i%7D%5E2%20%5Csin%5E2%28%5Ctheta_i%29%7Dg%20%5C%5C%5C%5C%20%5Cimplies%20R%20%3D%20%5Cdfrac%7B3%7Bv_i%7D%5E2%20%5Csin%5E2%28%5Ctheta_i%29%7Dg)
At the same time, the ball will have traveled half its horizontal range, so
![x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%20R2%20%3D%20v_i%20%5Ccos%28%5Ctheta_i%29%20%5Ctimes%20%5Cdfrac%7Bv_i%20%5Csin%28%5Ctheta_i%29%7Dg%20%5C%5C%5C%5C%20%5Cimplies%20R%20%3D%20%5Cdfrac%7B2%7Bv_i%7D%5E2%20%5Ccos%28%5Ctheta_i%29%20%5Csin%28%5Ctheta_i%29%7Dg)
Solve for
and
:
![\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7Bv_i%7D%5E2%20%5Csin%5E2%28%5Ctheta_i%29%7Dg%20%3D%20%5Cdfrac%7B2%7Bv_i%7D%5E2%20%5Ccos%28%5Ctheta_i%29%20%5Csin%28%5Ctheta_i%29%7Dg%20%5C%5C%5C%5C%20%5Cimplies%203%20%5Csin%5E2%28%5Ctheta_i%29%20%3D%202%20%5Ccos%28%5Ctheta_i%29%20%5Csin%28%5Ctheta_i%29%20%5C%5C%5C%5C%20%5Csin%28%5Ctheta_i%29%20%283%5Csin%28%5Ctheta_i%29%20-%202%20%5Ccos%28%5Ctheta_i%29%29%20%3D%200)
Since
, we cannot have
, so we're left with (e)
![3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}](https://tex.z-dn.net/?f=3%20%5Csin%28%5Ctheta_i%29%20-%202%5Ccos%28%5Ctheta_i%29%20%3D%200%20%5C%5C%5C%5C%20%5Cimplies%203%20%5Csin%28%5Ctheta_i%29%20%3D%202%5Ccos%28%5Ctheta_i%29%20%5C%5C%5C%5C%20%5Cimplies%20%5Ctan%28%5Ctheta_i%29%20%3D%20%5Cdfrac23%20%5C%5C%5C%5C%20%5Cimplies%20%5Cboxed%7B%5Ctheta_i%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%20%5Capprox%2033.7%5E%5Ccirc%7D)
Now,
![\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%5Cright%29%20%3D%20%5Cdfrac3%7B%5Csqrt%7B13%7D%7D)
![\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}](https://tex.z-dn.net/?f=%5Csin%5Cleft%28%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%5Cright%29%20%3D%20%5Cdfrac2%7B%5Csqrt%7B13%7D%7D)
so it follows that (d)
![R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}](https://tex.z-dn.net/?f=R%20%3D%20%5Cdfrac%7B2%7Bv_i%7D%5E2%20%5Ctimes%5Cfrac3%7B%5Csqrt%7B13%7D%7D%20%5Ctimes%20%5Cfrac2%7B%5Csqrt%7B13%7D%7D%7Dg%20%5C%5C%5C%5C%20%5Cimplies%20%7Bv_i%7D%5E2%20%3D%20%5Cdfrac%7B13Rg%7D%7B12%7D%20%5C%5C%5C%5C%20%5Cimplies%20%5Cboxed%7Bv_i%20%3D%20%5Csqrt%7B%5Cdfrac%7B13Rg%7D%7B12%7D%7D%7D)
Knowing the initial speed and angle, the initial vertical component of velocity is (c)
![v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}](https://tex.z-dn.net/?f=v_y%20%3D%20%5Csqrt%7B%5Cdfrac%7B13Rg%7D%7B12%7D%7D%20%5Csin%5Cleft%28%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac23%5Cright%29%5Cright%29%20%5C%5C%5C%5C%20%5Cimplies%20v_y%20%3D%20%5Csqrt%7B%5Cdfrac%7B13Rg%7D%7B12%7D%7D%20%5Ctimes%20%5Cdfrac2%7B%5Csqrt%7B13%7D%7D%20%5C%5C%5C%5C%20%5Cimplies%20%5Cboxed%7Bv_y%20%3D%20%5Csqrt%7B%5Cdfrac%7BRg%7D3%7D%7D)
We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)
![v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Csqrt%7B%5Cdfrac%7B13Rg%7D%7B12%7D%7D%20%5Ctimes%20%5Cdfrac3%7B%5Csqrt%7B13%7D%7D%20%5C%5C%5C%5C%20%5Cimplies%20v_x%20%3D%20%5Cdfrac%7B%5Csqrt%7B3Rg%7D%7D%7B2%7D)
Then with
, the ball's speed
is
![v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%7Bv_x%7D%5E2%20%2B%20%7Bv_y%7D%5E2%7D%20%5C%5C%5C%5C%20%5Cimplies%20v%20%3D%20v_x%20%5C%5C%5C%5C%20%5Cimplies%20%5Cboxed%7Bv%20%3D%20%5Cdfrac%7B%5Csqrt%7B3Rg%7D%7D2%7D)
Finally, in the work leading up to part (e), we showed the time to maximum height is
![t = \dfrac{v_i \sin(\theta_i)}g](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7Bv_i%20%5Csin%28%5Ctheta_i%29%7Dg)
but this is just half the total time the ball spends in the air. The total airtime is then
![2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}](https://tex.z-dn.net/?f=2t%20%3D%20%5Cdfrac%7B2%20%5Ctimes%20%5Csqrt%7B%5Cfrac%7B13Rg%7D%7B12%7D%7D%20%5Ctimes%20%5Cfrac2%7B%5Csqrt%7B13%7D%7D%7Dg%20%5C%5C%5C%5C%20%5Cimplies%202t%20%3D%202%5Csqrt%7B%5Cdfrac%20R%7B3g%7D%7D)
and the ball is in the air over the interval (a)
![\boxed{0 < t < 2\sqrt{\frac R{3g}}}](https://tex.z-dn.net/?f=%5Cboxed%7B0%20%3C%20t%20%3C%202%5Csqrt%7B%5Cfrac%20R%7B3g%7D%7D%7D)