Horizontally Launched Projectiles! Help!

A ball is thrown horizontally from the top of a building 18.2 m high. The ball strikes the ground at a point 83.2 m from the base of the building. The acceleration of gravity is 9.8 m/s.

a) Find the time the ball is in motion.
- I know the answer to this is 1.9262 but i cant figure out part b.

b) find the initial velocity of the ball.

At first i thought it was 0 but that's not right.

b) Vi = X/t = 83.2/1.9262 = 43.2 m/s

First, find the time it takes to hit the ground:

d = vi*t + 1/2*g*t^2 and since the vertical initial velocity is zero:

18.2 = .5*9.8*t^2 so t = 1.93 seconds

b) You are partially correct. The initial velocity in the y direction is zero,

but if the initial velocity in the x direction was also zero, the ball would drop

straight down and land right next to the building.

We will use d = r*t to find the rate in meters per second: 83.2 = r*1.93

so r which is vi in the x direction is 43.11 m/s

y = -g*x^2/(2*v^2)

=> v^2 = -g*x^2/(2*y)
=> v = sqrt(-g*x^2/(2*y)) =sqrt(-9.82*83.2^2/(2*(-18.2))) = 43 m/s.

I use eq. 2a from wiki (about midway down)

Initial velocity is the horizontal velocity = horizontal distance / travel time = 83.2/1.93 = 43.1 m/s
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