Question

A hot air balloon is ascending straight up at a constant speed of 8.40 m/s. When the balloon is 14.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 27.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places

Answer 1

Answer:

The objects have the same altitude at the same time in the heights 22.4m and 36.6m.

Explanation:

The equation of the position $y$ of the hot air balloon with respect to time is:

$y=y_0+v_{0B}t$

And the equation of the position $y$ of the pellet with respect to time is:

$y=v_{0P}t-\frac{1}{2} gt^{2}$

As the altitude of the two objects must be equal, we can match these equations, so:

$y_{0B}+v_{0B}t=v_{0P}t-\frac{1}{2} gt^{2} \\\\\frac{1}{2} gt^{2}+(v_{0B}-v_{0P})t+y_{0B}=0\\\\4.9\frac{m}{s^{2}}t^{2} -18.6\frac{m}{s} t+14.0m=0$

Solving this equation for t, we obtain:

$t=\frac{18.6\frac{m}{s} \±\sqrt{345.9\frac{m^{2}}{s^{2}}-4(4.9\frac{m}{s^{2}})(14.0m)}}{9.8\frac{m}{s^{2}}}=\frac{18.6\frac{m}{s}\±8.4\frac{m}{s}}{9.8\frac{m}{s^{2}}} \\\\\implies \left \{ {{t_1=1.0s} \atop {t_2=2.7s}} \right.$

Now, using the equation of the position of the balloon (we can use the other either, but the first one is easier) we can obtain the height of the places where the two objects have the same altitude at the same time:

$y_1=14.0m+8.40\frac{m}{s} (1.0s)=22.4m\\\\y_2=14.0m+8.4\frac{m}{s}(2.7s)=36.6m$

In words, the hot air balloon and the pellet have the same altitude at the same time in the heights 22.4m and 36.6m.