Imagine the ball on the left is given a nonzero initial speed in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. what horizontal speed v x must the ball on the left start with so that it hits the ground at the same position as the ball on the right? remember that when the two balls are released, they are starting at a horizontal distance of 3.0 m apart.

The horizontal speed of a projectile is constant

The required starting horizontal speed of the ball on the left to hit the ground at the same position as the ball on the right, vₓ is 3.0 m/s

Question: Parts of the question missing are;

The height from which the balls are dropped, h = 5.0 m

The acceleration due to gravity, g ≈ 10.0 m/s²

The attached diagram

The known parameter;

The distance between the two balls = 3.0 m

Solution:

The time it will take the ball to fall from the 3.0 meters height, is given by the equation of free fall as follows;

h = (1/2)·g·t²

Which gives;

t = √((2·h)/g)

Therefore;

t = √((2×5)/10) = 1.0

The time it takes the ball to fall from the 5.0 meters, height, t = 1.0 seconds

The time it will take the balls to reach the ground, t = 1.0 s = The time the ball on the left has to travel the d = 3.0 m horizontal distance apart from which the two balls start

Therefore;

The horizontal speed the ball on the left has to start with so that it hits the ground at the same position as the ball on the right, vₓ = 3.0 m/s

Learn more about projectile motion here:

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Let say the height of two balls from the ground is H

now we can use kinematics

now we have

now in the same time ball on the left will cover the horizontal distance between them

so above is the horizontal speed of the left ball