A bullet of mass 25 g is fired horizontally into a 5.0 kg ballistic...

Let $m$ be the mass of the bullet, $M$ be the mass of the pendulum, $v$ be the initial speed of the bullet, and $V$ be the speed of the combined mass just after the collision.

By conservation of linear momentum:

$m v = (m + M) V$

$V = \dfrac{m v}{m + M}$

By conservation of mechanical energy after the collision, as the pendulum rises to a height $h$:

$\dfrac{1}{2} (m + M) V^2 = (m + M) g h$

$V = \sqrt{2 g h}$

Equating the two expressions for $V$:

$\dfrac{m v}{m + M} = \sqrt{2 g h}$

$v = \left( \dfrac{m + M}{m} \right) \sqrt{2 g h}$

Given $m = 25 \text{ g} = 0.025 \text{ kg}$, $M = 5.0 \text{ kg}$, $h = 10 \text{ cm} = 0.1 \text{ m}$, and taking $g = 9.8 \text{ m/s}^2$:

$v = \left( \dfrac{0.025 + 5.0}{0.025} \right) \sqrt{2 \times 9.8 \times 0.1}$

$v = \left( \dfrac{5.025}{0.025} \right) \sqrt{1.96}$

$v = 201 \times 1.4 = 281.4 \text{ m/s}$

Approximating $m + M \approx M$ (since $m \ll M$):

$v \approx \dfrac{5.0}{0.025} \times 1.4 = 200 \times 1.4 = 280 \text{ m/s}$