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
As described above, the acceleration of the object is given by the equation, a=2d/t^2. If the object accelerates at a distance of 15 meters, then it can be said that the acceleration will be doubled as compared to the previous with the same time squared. Time can be related by the equation,
t=sqrt of 30/a
For the same wave, the product product of
(wavelength) times (frequency)
is always the same number. (It happens to be the speed of the wave.)
So if one of them changes, the other one has to change in the opposite
direction, in order to keep their product constant.
For electromagnetic waves, higher frequency means higher energy.
I'm not sure about mechanical waves just now.