A stone is thrown vertically downward with an initial speed of 12.0 m/s from the top of a building. The stone takes 1.54 s to reach the ground. a) What is the height of the building?
b) What is the speed of the stone when it hits the ground?

(a) The stone travels a vertical distance y of

y = (12.0 m/s) t + 1/2 g t ²

where g = 9.80 m/s² is the acceleration due to gravity. Note that this equation assume the downward direction to be positive, and that y = 0 corresponds to the height from which the stone is thrown.

So if it reaches the ground in t = 1.54 s, then the height of the building y is

y = (12.0 m/s) (1.54 s) + 1/2 (9.80 m/s²) (1.54 s)² ≈ 30.1 m

(b) The stone's (downward) velocity v at time t is

v = 12.0 m/s + g t

so that after t = 1.54 s, its velocity is

v = 12.0 m/s + (9.80 m/s²) (1.54 s) ≈ 27.1 m/s

(and of course, speed is the magnitude of velocity)

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

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