Answer
The second projectile was faster than the first projectile by a factor of $\sqrt{2}$
Work Step by Step
Let $m$ be the mass of a projectile and let $M$ be the mass of the pendulum. Let $v$ be the initial speed of a projectile.
We can use conservation of momentum to find the speed $v'$ just after the collision:
$m~v = (m+M)~v'$
We can use conservation of energy to find the height $h$ reached by the pendulum as it swings up:
$\frac{1}{2}(m+M)~(v')^2 = (m+M)~gh$
$v' = \sqrt{2gh}$
We can replace $v'$ in the first equation above:
$m~v = (m+M)\sqrt{2gh}$
$v = \frac{m+M}{m}\sqrt{2gh}$
Let $v_2$ be the initial speed of the second projectile:
Let $v_1$ be the initial speed of the first projectile.
$v_2 = \frac{m+M}{m}\sqrt{2g(2h)}$
$v_1 = \frac{m+M}{m}\sqrt{2g(h)}$
We can divide the $v_2$ equation by the $v_1$ equation:
$\frac{v_2}{v_1} = \sqrt{\frac{2h}{h}} = \sqrt{2}$
$v_2 = \sqrt{2}\times v_1$
The second projectile was faster than the first projectile by a factor of $\sqrt{2}$
