A man is riding a flatcar traveling with constant speed of 4 m/s. There is a stationary ring about the line of motion of the car. Height of the ring above the man’s hand is 7.2 m. He wishes to throw a ball through the ring in such a manner that the ball passes through the ring moving horizontally. He throws the ball with the speed of 20 m/s with respect to himself. At what horizontal distance in front of the ring must he release the ball?
According to my calculations (which have been verified), the vertical component of the velocity is 12m/s since $$7.2=\frac{u^2\sin^2\theta}{2g}$$ $$u\sin\theta=12$$ Since u is given as 20, angle of velocity with respect to the car is $$\theta=37^\circ$$
The time taken for the ball to reach the ring will be $$T=\frac{u\sin\theta}{g}=1.2s$$
Now we need to find the horizontal distance before the ring. Since it moves horizontally in the ring, that distance is half the range $$\frac R2=\frac{u^2\sin2\theta}{2g}$$$$=\frac{400\times 2 \times 3 \times 4}{10\times 5 \times 5 \times 2}$$ $$=19.2m$$
The right answer is 24m. What am I doing wrong?
You found the horizontal distance the ball will travel from the thrower's hand to the top of its arc in the frame of reference attached to the flatbed car. In the frame of reference of the ground, the horizontal component of the ball's velocity is $4\ \mathrm{m/s}$ greater than the horizontal component in the flatbed car's frame of reference. The distance traveled relative to the ground is how far the thrower must be from the ring (measuring only the horizontal distance).
Alternatively, if you are comfortable working in the flatbed car's frame of reference, the distance to the top of the arc is $19.2\ \mathrm m,$ as you calculated, but at the instant when the ball is released the ring is not there. The ring is approaching from some distance further and must travel for $1.2$ seconds at $4\ \mathrm{m/s}$ in order to reach the place where the ball passes through it.