1. Geometric Setup & Tangency Condition

A particle is projected horizontally with speed $u$ from height $h$ directly above the center of a hemisphere of radius $R = 10\sqrt{2}$ m. The trajectory is a parabola that must touch (be tangent to) the hemisphere to ensure minimum energy while clearing it.

Equation of Trajectory: $y = h – \frac{gx^2}{2u^2}$

Equation of Hemisphere: $x^2 + y^2 = R^2$

Substituting $x^2 = R^2 – y^2$ into the trajectory equation:

$$ y = h – \frac{g(R^2 – y^2)}{2u^2} \implies 2u^2y = 2u^2h – gR^2 + gy^2 $$ $$ gy^2 – 2u^2y + (2u^2h – gR^2) = 0 $$

For the path to touch the dome, this quadratic in $y$ must have a single root (Discriminant $D=0$):

$$ D = (2u^2)^2 – 4g(2u^2h – gR^2) = 0 $$ $$ u^4 – 2ghu^2 + g^2R^2 = 0 $$ Solving for $u^2$: $$ u^2 = g(h – \sqrt{h^2 – R^2}) $$ (We choose the negative root to minimize velocity).

2. Minimizing Ground Speed

The speed at the ground ($v_g$) is given by energy conservation: $v_g^2 = u^2 + 2gh$.

Substituting $u^2$: $$ v_g^2 = g(h – \sqrt{h^2 – R^2}) + 2gh = g(3h – \sqrt{h^2 – R^2}) $$ To find minimum speed, we minimize $f(h) = 3h – \sqrt{h^2 – R^2}$. $$ \frac{df}{dh} = 3 – \frac{h}{\sqrt{h^2 – R^2}} = 0 $$ $$ 3\sqrt{h^2 – R^2} = h \implies 9(h^2 – R^2) = h^2 \implies 8h^2 = 9R^2 $$ $$ h = \frac{3R}{2\sqrt{2}} $$ Given $R = 10\sqrt{2}$ m: $$ h = \frac{3(10\sqrt{2})}{2\sqrt{2}} = 15 \text{ m} $$

3. Calculating Values

Height: $h = 15$ m.

Projection Velocity ($u$):

$$ u^2 = g(h – \sqrt{h^2 – R^2}) $$ At $h = \frac{3R}{2\sqrt{2}}$, we have $\sqrt{h^2 – R^2} = \frac{R}{2\sqrt{2}}$. $$ u^2 = g\left( \frac{3R}{2\sqrt{2}} – \frac{R}{2\sqrt{2}} \right) = \frac{gR}{\sqrt{2}} $$ $$ u = \sqrt{\frac{10(10\sqrt{2})}{\sqrt{2}}} = \sqrt{100} = 10 \text{ m/s} $$

Minimum Ground Speed ($v_g$):

$$ v_g = \sqrt{u^2 + 2gh} = \sqrt{100 + 2(10)(15)} = \sqrt{400} = 20 \text{ m/s} $$