Answer
28 m/s
Work Step by Step
Let's take,
Initial velocity of the car relative to the road = $V_{CR}$
Velocity of the car relative to the first truck = $V_{CT}$
Velocity of the first truck relative to the road = $V_{TR}$
Let's apply equation 3.5a $S=ut$ in the horizontal direction to the car to find the flight time.
$\rightarrow S=ut$ ; Let's plug known values into this equation.
$15\space m=V_{CT}cos16^{\circ}t=>t=\frac{15\space m}{V_{CT}cos16^{\circ}}$
Let's apply equation 3.5b $S=ut+\frac{1}{2}at^{2}$ in the vertical direction of the car to find the velocity relative to the first truck.
$\uparrow S=ut+\frac{1}{2}at^{2}$ ; Let's plug known values into this equation.
$0=V_{CT}sin16^{\circ}t+\frac{1}{2}(-9.8\space m/s^{2})t^{2}$
$V_{CT}=\frac{4.9\space m/s^{2}}{sin16^{\circ}}\times \frac{15\space m}{V_{CT}cos16^{\circ}}$
$V_{CT}=\sqrt {\frac{4.9\times15}{cos16^{\circ}sin16^{\circ}}}\approx 17\space m/s$
We can write,
$V_{CR}=V_{CT}+C_{TR}=17\space m/s+11\space m/s=28\space m/s$
