Aball is thrown horizontally from the top of a 65 m building and lands 115 m from the base of the building. ignore air resistance, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.
(a) how long is the ball in the air in seconds?
(b) what must have been the initial horizontal component of the velocity in m/s?
(c) what is the vertical component of the velocity just before the ball hits the ground in m/s?
(d) what is the magnitude of the velocity of the ball just before it hits the ground in m/s?

The answer to your question is:

a) t = 3.64 s

b) vox = 31.59 m/s

c) vy = 35.71 m/s

d) v = 47.67 m/s

Explanation:

a) To calculate the time, we know that voy = 0 m/s so we this this formula

   h = voy + 1/2(gt²)

   65  = 0 + 1/2(9.81)(t²)

   65 = 4.905t²

   t² = 65/4.905 = 13.25

   t = 3.64 s

b) To calculate vox  we use this formula vox = d/t ; vox is constant

   vox = 115/3.64 = 31.59 m/s

c) To calculate voy we use the formula    vy = voy + gt

but voy = 0

               vy = gt = 9.81 x 3.64 = 35.71 m/s

d) To calculate v  we use the pythagorean theorem

            c2 = a2 + b2

            c2 = 31.59² + 35.71² = 997.92 + 1275.20

            c2 = 2273.12

              c = 47.67 m/s

do you have a formula possibly