Before Being Thrown - Speed is zero (0). This is your answer.
After Being Thrown - Try that again. Look for linear elements.
Could someone take a look at my answers?
After being thrown from the top of a tall building, a projectile follows a path described parametrically by (x, y) = (48t, 400 − 16t2), where x and y are in feet and t is in seconds.
(a) How many seconds did it take for the object to reach the ground, where
y=0? How far from the building did the projectile land?
(b) How fast was the projectile moving at t = 0 when it was thrown?
(c) Where was the projectile when t = 2, and (approximately) how fast was it moving?
If the ball were thrown with speed 0, it would fall straight down. That does not happen. I do note that the -16t^2 is due to the acceleration of gravity and the 400 is the height of the building so the y- component is that for "falling". But the horizontal position is given by 48t so, for example, it will move 48 feet in one second. What speed is that?