Answer:
[tex]t=5.30s[/tex]
Explanation:
The high reached by a proyectile in an uniformly accelerated motion is given by:
[tex]y=v_{0y}t-\frac{gt^2}{2}[/tex]
The time that the rocket spends in the air is obtained for y = 0, since this is the time that the rocket travels before touching the ground. Recall that [tex]v_{0y}=v_0sin\theta[/tex]. Solving for t:
[tex]0=(v_0sin\theta) t-\frac{gt^2}{2}\\\frac{gt}{2}=v_0sin\theta\\t=\frac{2v_0sin\theta}{g}\\t=\frac{2(30\frac{m}{s})sin(60^\circ)}{9.8\frac{m}{s^2}}\\t=5.30s[/tex]
