Let the origin be the point on the ground directly below the airplane at the moment the bomb is released ($t=0$). The airplane's velocity is denoted by $v_p$. The problem is analyzed by separating the motion into vertical and horizontal components.

1. Time of Flight The bomb is released from a height $h$ with zero initial vertical velocity. Its vertical motion is governed by gravity. The time of flight, $t_f$, is the time it takes to reach the ground ($y=0$). Using the kinematic equation for vertical displacement:

$$y(t) = y_0 + v_{y0}t - \frac{1}{2}gt^2$$

With $y(t_f) = 0$, $y_0 = h$, and $v_{y0} = 0$:

$$0 = h - \frac{1}{2}gt_f^2$$

Solving for the time of flight:

$$t_f = \sqrt{\frac{2h}{g}}$$

2. Horizontal Motion and Intercept Condition For the bomb to hit the target, both must have the same horizontal position at $t = t_f$. The bomb's initial horizontal velocity is the airplane's velocity, $v_p$. Its horizontal position is:

$$x_b(t) = v_p t$$

The target starts at a horizontal distance $d$ and moves with speed $v_t$. Its horizontal position is:

$$x_t(t) = d + v_t t$$

Setting $x_b(t_f) = x_t(t_f)$ for a successful hit:

$$v_p t_f = d + v_t t_f$$

3. Derivation of Airplane Speed ($v_p$) Rearranging the intercept equation to solve for $v_p$:

$$v_p t_f - v_t t_f = d$$ $$(v_p - v_t) t_f = d$$ $$v_p = v_t + \frac{d}{t_f}$$

Substitute the expression for the time of flight, $t_f$:

$$v_p = v_t + \frac{d}{\sqrt{\frac{2h}{g}}}$$

This simplifies to the final expression for the airplane's speed.