Let the initial temperature be $t$, the temperature after the fall be $t_{1}$, the mass of the lead $m$, and the path of the falling body $s$.

It is necessary that the lost potential energy be equal to the heat gain:

$$
m g s=\left(t_{1}-t\right) m E C
$$

from which

$$
t_{1}=t+\frac{g s}{E C}=93.2^{\circ}
$$

If in the second case we want the heat gain to be greater, the body must be given an initial velocity $v_{0}$; in this case:

$$
m g s+\frac{m v_{0}^{2}}{2}=(T-t) m E C
$$

from which

$$
v_{0}=\sqrt{2[(T-t) E C-g s]}=80.46 \frac{\mathrm{m}}{\mathrm{sec}}
$$

(Riesz Frigyes, Gymn. VIII. o. t., Györ.)

The problem was also solved by: Szabó István.
A lead bullet at $20^{\circ}$ temperature, completely inelastic, falls freely from a height of $100 \mathrm{~m}$ onto a solid plane. If all the apparently lost energy was converted into heat absorbed by the bullet, the questions are:

1. What is the temperature of the bullet immediately after the fall?
2. What speed would the bullet need to be dropped from above to raise its temperature to the melting point of lead?

The specific heat of lead $C=0.0315$, the mechanical equivalent of heat $E=425$, the melting point of lead $T=335^{\circ}$, and the intensity of gravitational force $g=9.80 \frac{\mathrm{m}}{\mathrm{s}^{2}}$.