Question

If a horizontally projected object falls a vertical distance of 1.000 m and its range is 1.000 m, what is its initial horizontal velocity?

Answer 1

Initial vertical velocity=0=u_y
Vertical distance=1 m=s_y

\[S_y=u_yt-\frac{1}{2}gt^2\]
1=\(\frac{1}{2}gt^2\)

Find t from here

Answer 2

Range = horizontal velocity*time
Horizontal velocity will remain same throughout the motion.

1=\(u_x*t\)

Answer 3

Find u_x which is horizontal velocity