A snowball is launched horizontally from the top of a building at v = 19.6 m/s. If it lands d = 64 meters from the bottom, how high (in m) was the building?

height of building = 18.8m

Explanation: The question is a projectile motion, a two dimensional motion with a vertical constant acceleration (g = - 9.8m/s²) and a constant horizontal velocity (thus making horizontal component of acceleration zero).

From the question, distance between bottom of building and where the object lands = 64m, initial velocity for throwing the object = 19.6m/s

The horizontal range formulae is given as

d= vt

Where d= horizontal range = 64m, v = initial velocity of throw.

64 = 19.6 × t

t = 64/ 19.6

t = 3.265 s.

Height (h) of the building is gotten by using the formulae

h =vt - 1/2gt²

h = (19.6×3.265) - 1/2×9.8×(3.265)²

h = 71.05 - (104.47/2)

h = 71.05 - 52.235

h = 18.8m

the body's situation won't change the forces are equal opposit and on the same body

it breaks and dirupture other materials and other sencews from the bottom