Answer
$9.0~m/s$
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.
$x_{0_c}=0$, $v_c=2.0~m/s$, $a=-g=-9.81~m/s^2$
$v_c=v_{0_c}+at$
$2.0~m/s=v_{0_c}+(-9.81~m/s^2)t$
$(9.81~m/s^2)t+2.0~m/s=v_{0_c}$
$x_c=x_{0_c}+v_{0_c}t+\frac{1}{2}at^2$
$x_c=0+[(9.81~m/s^2)t+2.0~m/s]t+\frac{1}{2}(-9.81~m/s^2)t^2$
$x_c=(2.0~m/s)t+(4.905~m/s^2)t^2$
When the camera reaches the passenger:
$x_c=x_p$
$(2.0~m/s)t+(4.905~m/s^2)t^2=2.5~m+(2.0~m/s)t$
$(4.905~m/s^2)t^2=(2.5~m)$
$t=±\sqrt{\frac{2.5~m}{4.905~m/s^2}}$. The valid solulution is:
$t=0.714~s$
$v_{0_c}=(9.81~m/s^2)t+2.0~m/s$. Make $t=0.714~s$:
$v_{0_c}=(9.81~m/s^2)(0.714~s)+2.0~m/s$
$v_{0_c}=9.0~m/s$
