A solid cylinder having radius R and length L is slipping on a...

Initial translational velocity, $v_0 = 49$ m/s.

Initial angular velocity, $\omega_0 = \dfrac{v_0}{4R}$.

Velocity of the lowest point of the cylinder is $v_{bottom} = v_0 - \omega_0 R = v_0 - \dfrac{v_0}{4} = \dfrac{3v_0}{4} > 0$.

Since the lowest point is slipping forward, kinetic friction $f_k = \mu_k m g$ acts in the backward direction.

Translational acceleration, $a = -\dfrac{f_k}{m} = -\mu_k g$.

Angular acceleration, $\alpha = \dfrac{f_k R}{I} = \dfrac{\mu_k m g R}{\dfrac{1}{2} m R^2} = \dfrac{2\mu_k g}{R}$.

At time $t$, the translational velocity is $v = v_0 - \mu_k g t$.

The angular velocity is $\omega = \omega_0 + \alpha t = \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t$.

For pure rolling to start, $v = \omega R$.

$v_0 - \mu_k g t = \left( \dfrac{v_0}{4R} + \dfrac{2\mu_k g}{R} t \right) R$

$v_0 - \mu_k g t = \dfrac{v_0}{4} + 2\mu_k g t$

$3\mu_k g t = \dfrac{3v_0}{4}$

$t = \dfrac{v_0}{4\mu_k g}$

Substituting the given values $v_0 = 49$ m/s, $\mu_k = 0.25$, and $g = 9.8$ m/s$^2$:

$t = \dfrac{49}{4 \times 0.25 \times 9.8} = \dfrac{49}{9.8} = 5$ s.

Answer: $5$