Determine the horizontal range, d of the ball in terms of uh and h. What is the proportionality constant?

Question

A ball is thrown horizontally with an initially speed u(small h) from the top of a building of height h. Find that d opposes Squareroot h

Answer 1

Basically, since gravity acts at ~9.8 meters per second squared, the downward aspect of the velocity of the ball at any time is equal to ~9.8 multiplied by the time (in seconds). This works through integration: 9.8 -> 9.8x -> 4.9x^2. So the distance traveled vertically varies as the square of the time, and since the horizontal velocity does not change (ignoring air resistance) over time, it varies linearly as the time. Therefore, the vertical drop varies as the square of the horizontal distance d travelled in a certain amount of time. Say we let t = the number of seconds since the ball was thrown, the vertical drop 'h' = 4.9 * t^2, whereas the horizontal distance 'd' = uh * t. Therefore t = d/uh. Substituting that into h, we get h=4.9*d^2/uh^2. if we let 4.9/uh^2=k, we get h =k*d^2, and so h varies as the square of d. Squareroot of both sides means d varies as the squareroot h. Sorry if it was poorly explained and set out.