Answer: 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
