You mount a ball launcher to a car. It points toward the back of the car 39 degrees above the horizontal and launches balloons at 9.7 m/s. When you drive the car straight ahead at 2.2 m/s. Assume the ball is launched from ground level. For the ball, what is the separation between the ball and the car in m when the ball hits the ground?

We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.

If the ballon is launched at 9.7 m/s at 39 degrees of elevation.

Vx0 = 9.7 * cos(39) = 7.5 m/s

Vy0 = 9.7 * sin(39) = 6.1 m/s

If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 = 0

a = -9.81 m/s^2

It will fall when Y(t) = 0

0 = 6.1 * t - 4.9 * t^2

0 = t * (6.1 - 4.9 * t)

t1 = 0 (this is when the balloon was launched)

0 = 6.1 - 4.9 * t2

4.9 * t2 = 6.1

t2 = 6.1 / 4.9 = 1.25 s

The distance from the car will be the horizonta distance it travelled in that time

X(t) = X0 + Vx0 * t

X(1.25) = 7.5 * 1.25 = 9.4 m