Question

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.

Answer 1

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

now we can use kinematics

$s = v_i * t + \frac{1}{2} at^2$

now we have

$H = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2H}{g}}$

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

$v_x = \frac{d}{ t}[/tex[tex]v_x = \frac{3}{\sqrt{\frac{2H}{g}}}$

so above is the horizontal speed of the left ball