a ball dropped from the roof of a building takes 4s to reach the street. how high is the building?
The distance travelled (d) by an accelerating body with initial velocity u and acceleration a is obtainable from the equation of motion of the form:
`d = ut + 1/2at^2`
Here, the ball, initially at rest, is being dropped from a height, h. So, u = 0, a = g = 9.81 m/s^2, t = 4 s
Putting the values we get...
`h = 0*4 + (1/2) *9.81*4^2`
= `78.48` m
= 78.5 m (rounded off to one decimal place).
Therefore, the height of the building was 78.5 m.