Answer:3.085 m ,5.914 m
Explanation:
Given
Velocity of hot air balloon(v)=2 m/s
Velocity at which camera is thrown up(u)=12 m/s
at t=0 passenger is at 2.5 m above
Let say it takes t sec for camera to reach balloon and balloon has traveled a distance of x m in that time
x=2t
also for camera
[tex]2.5+x=12\times t-\frac{1}{2}gt^2[/tex]------1
substitute value of x in 1
[tex]2.5+2t=12t-5t^2[/tex]
[tex]5t^2-10t+2.5=0[/tex]
[tex]t^2-2t+0.5=0[/tex]
[tex]2t^2-4t+1=0[/tex]
[tex]t=\frac{4\pm \sqrt{8}}{4}=\frac{2\pm \sqrt{2}}{2}[/tex]
so there are two instances when both heights are equal
for [tex]t=\frac{2-\sqrt{2}}{2}[/tex]
[tex]x=2\times \frac{2-\sqrt{2}}{2}[/tex]
[tex]x=2-\sqrt{2}=0.5857 m[/tex]
Therefore balloon is at a height of 2.5+0.5857=3.085 m
For [tex]t=\frac{2+\sqrt{2}}{2}[/tex]
[tex]x=2+\sqrt{2}[/tex]
balloon at a height of 5.914 m
