Answer
$3.0~m$
Work Step by Step
Let the positive direction be upward with origin at the initial position of the camera.
- The passenger: constant velocity.
$x_{0_p}=2.5~m$, $v_p=2.0~m/s$
$x_p=x_{0_p}+v_pt$
$x_p=2.5~m+(2.0~m/s)t$
- The camera: constant acceleration.
$x_{0_c}=0$, $v_{0_c}=13~m/s$, $a=-g=-9.81~m/s^2$
$x_c=x_{0_c}+v_{0_c}t+\frac{1}{2}at^2$
$x_c=0+(13~m/s)t+\frac{1}{2}(-9.81~m/s^2)t^2$
$x_c=(13~m/s)t-(4.905~m/s^2)t^2$
When the camera reaches the passenger:
$x_p=x_c$
$2.5~m+(2.0~m/s)t=(13~m/s)t-(4.905~m/s^2)t^2$
$(4.905~m/s^2)t^2-(11~m/s)t+(2.5~m)=0$
This is a quadratic equation for $t$.
$t=\frac{-(-11~m/s)±\sqrt {(-11~m/s)^2-4(4.905~m/s^2)(2.5~m)}}{2(4.905~m/s^2)}$
$t_1=1.99~s$
$t_2=0.257~s$
Interpretation: the camera is launched with an initial speed of $13~m/s$. When the camera reaches the passenger, it is moving upward, faster than the ballon. Suppose the passenger does not catch the camera: the camera will pass the balloon, reach its maximum height and come back. Then it will pass the balloon a second time, but, now, moving downward.
We are assuming the passenger will catch the camera when it reaches the ballon the first time.
$t=0.257~s$.
$x_p=2.5~m+(2.0~m/s)t$. Now, make $t=0.257~s$
$x_p=2.5~m+(2.0~m/s)(0.257~s)=3.0~m$
