Originally Posted by

**free_to_fly** I would like some help on this question please, just spent the last hour trying to figure it out.

A smooth hemispherical bowl of radius a has its rim horizontal. A particle P os projected downwards with speed u from a point on the surface of the bowl 0.5a below the level of the rim. If P remains in contact with the bowl during the subsquent motion, show that $\displaystyle \

u \le \sqrt {ag}

\$

Let $\displaystyle \

R_1

\$ be the reaction force from the point of projection.

$\displaystyle \

R_2

\$ be the reaction force at the lowest point, eg bottom of the bowl.

Let v be the speed of P when passing the lowest point.

I've used PE lost=KE gained and F=ma to come up with 3 equations:

$\displaystyle \

\begin{array}{l}

ag = v^2 - u^2 \\

R_1 = \frac{{mu^2 }}{a} + \frac{{mg}}{2} \\

R_2 = \frac{{mv^2 }}{a} + mg \\

\end{array}

\$

But I'm not sure what to do after this. Any help would be loved.