Answer:
[tex]v_0=9.9\ m.s^{-1}[/tex]
Explanation:
Given:
We have formula for the range of projectile:
[tex]R=\frac{v_0^2\times sin\ 2\theta}{g}[/tex]
putting the respective values
[tex]5=\frac{v_0^2\times sin\ 30^{\circ}}{9.8}[/tex]
[tex]v_0=9.9\ m.s^{-1}[/tex] is the velocity with which Tom should jump to land on the other roof.