Answer:
The balloon is 66.62 m high
Explanation:
Combined Motion
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is
[tex]\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}[/tex]
The values are
[tex]\displaystyle y_o=15\ m[/tex]
[tex]\displaystyle v_o=14\ m/s[/tex]
We must find the values of t such that the height of the camera is 0 (when it hits the ground)
[tex]\displaystyle y=0[/tex]
[tex]\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0[/tex]
Multiplying by 2
[tex]\displaystyle 2y_o+2v_ot-gt^2=0[/tex]
Clearing the coefficient of [tex]t^2[/tex]
[tex]\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0[/tex]
Plugging in the given values, we reach to a second-degree equation
[tex]\displaystyle t^2-2.857t-3.061=0[/tex]
The equation has two roots, but we only keep the positive root
[tex]\displaystyle \boxed {t=3.69\ s}[/tex]
Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is
[tex]\displaystyle Y_r=15+V_r.t[/tex]
[tex]\displaystyle Y_r=15+14\times3.69[/tex]
[tex]\displaystyle \boxed{Y_r=66.62\ m}[/tex]
