Respuesta :
Answer:t=0.3253 s
Explanation:
Given
speed of balloon is [tex]u=3\ m/s[/tex]
speed of camera [tex]u_1=20\ m/s[/tex]
Initial separation between camera and balloon is [tex]d_o=5\ m[/tex]
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is
[tex]s=ut[/tex]
[tex]s=3\times t[/tex]
and distance travel by camera to reach balloon is
[tex]s_1=ut+\frac{1}{2}at^2[/tex]
[tex]s_1=20\times t-\frac{1}{2}gt^2[/tex]
Now
[tex]\Rightarrow s_1=5+s[/tex]
[tex]\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t[/tex]
[tex]\Rightarrow 5t^2-17t+5=0[/tex]
[tex]\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}[/tex]
[tex]\Rightarrow t=\dfrac{17\pm 13.747}{10}[/tex]
[tex]\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s[/tex]
There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is [tex]t=0.3253\ s[/tex]
velocity is given by
[tex]v=u+at[/tex]
[tex]v=20-10\times 0.3253[/tex]
[tex]v=16.747\ m/s[/tex]
and position of camera is same as of balloon so
Position is [tex]=5+3\times 0.3253[/tex]
[tex]=5.975\approx 6\ m[/tex]
