Hit The Right Angle

Equation of trajectory is -

$\large y = x\tan\theta - \frac{gx^2}{2V_0 ^2cos^2 \theta}$

$\large y = - \frac{gx^2}{2V_0 ^2}$

$\large \frac{dy}{dx} = -\frac{gx}{V_0 ^2}$

This is trajectory is perpendicular to planck when it hits it.

$\therefore m_1m_2 = -1$

$\large x = \frac{\sqrt{3} V_0 ^2}{g}$

Also

$x = V_0 t$

$\large \therefore t = \sqrt{3} \frac{V_0 }{g}$

Now, $\large \tan(60) = \sqrt(3) = \frac{V_0 t}{H - 0.5gt^2}$

Equating, we get-

$\large \frac{\sqrt{3}gt^2}{2} + V_0 t - \sqrt{3}H = 0$

Plugging in the value of $t$ , we get-

$\large \frac{2gH}{V_0 ^2} = 5$

Suppose the particle covered distance 's' vertically before striking the plank. Then 2gs=(√3 V°)^2. AS velocity in vertical direction at the time of striking the plank is √3V° as velocity at time of striking made angle 30° with vertical and 60° with horizontal by geometry. Therefore horizontal distance travelled by particle before striking is V°^2√3/g. s=3V°^2/2g. H—s = V°^2/g by trignometry. Therefore H = 5V°^2/2g. So, 2gH/ V°^2 = (5.).

The velocity of the projectile over time is given by $\vec{v} = \langle v_0, -gt\rangle$ and the direction of the hill is given by $\vec{h} = \langle \cos \pi/6, \sin \pi/6\rangle$ . The projectile collides with the hill perpendicularly if the dot product of its velocity with the direction of the hill is zero. This leads us to the time of collision with the wall a la $\begin{aligned} \vec{v}\cdot\vec{h} &= 0 \\ &= v_0\cos\pi/6 - gt\sin\pi/6 \\ &\rightarrow t = \frac{v_0}{g}\cot\pi/6 \end{aligned}$ The height of the cliff is then the height the projectile falls before colliding with the hill $h_1$ , plus the height of the hill at which it hits $h_2$ . By trigonometry, we see that $\tan\pi/6 = h_2/r_x$ where $r_x$ is the $x$ position at which it hits the hill. Therefore, the total height $H = h_1 + h_2$ is given by $\begin{aligned} H &= \overbrace{r_x}^{v_0t}\tan\pi/6 + \frac12 gt^2 \\ &= \frac{v_0\tan\pi/6 v_0 \cot\pi/6}{g} + \frac12 g\frac{v_0^2\cot^2\pi/6}{g^2} \\ &= \frac{v_0^2}{g}\left(1 + \frac32\right)\\ &\rightarrow \frac{2gH}{v_0^2} = 5 \end{aligned}$