We can solve this problem by applying the principle of conservation of mechanical energy, as the only force doing work is gravity. Let the initial velocity of the object be $v_0$. Since the object is projected horizontally, this is its initial speed.

The mechanical energy of the object is the sum of its kinetic energy ($K$) and gravitational potential energy ($U$).

At the initial point (height $h$, speed $v_0$): The initial kinetic energy is $K_i = \frac{1}{2}mv_0^2$. The initial potential energy is $U_i = mgh$. The total initial energy is $E_i = K_i + U_i = \frac{1}{2}mv_0^2 + mgh$.

At the final point (on the ground, height 0, speed $v$): The final kinetic energy is $K_f = \frac{1}{2}mv^2$. The final potential energy is $U_f = 0$. The total final energy is $E_f = K_f + U_f = \frac{1}{2}mv^2$.

According to the conservation of mechanical energy, $E_i = E_f$:

$$\frac{1}{2}mv_0^2 + mgh = \frac{1}{2}mv^2$$

To find the initial velocity $v_0$, we solve this equation algebraically. We can cancel the mass $m$ from each term:

$$\frac{1}{2}v_0^2 + gh = \frac{1}{2}v^2$$

Now, we isolate $v_0^2$:

$$v_0^2 = v^2 - 2gh$$

Finally, taking the square root gives the magnitude of the initial velocity:

$$v_0 = \sqrt{v^2 - 2gh}$$