Question

A student launches a small 0.5 kg rocket with an initial speed of 30 m/s at an angle of 60°. Approximately how much time will the rocket spend in the air? Air resistance is negligible.

Answer 1

Answer:

$t=5.30s$

Explanation:

The high reached by a proyectile in an uniformly accelerated motion is given by:

$y=v_{0y}t-\frac{gt^2}{2}$

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 $v_{0y}=v_0sin\theta$. Solving for t:

$0=(v_0sin\theta) t-\frac{gt^2}{2}\\\frac{gt}{2}=v_0sin\theta\\t=\frac{2v_0sin\theta}{g}\\t=\frac{2(30\frac{m}{s})sin(60^\circ)}{9.8\frac{m}{s^2}}\\t=5.30s$