A ball is thrown downward from the top of a building with an initial speed of 25m/s. it strikes the ground after 2.0s. how high is the building?

For this case we have an equation of the form:

h (t) = (1/2) * (a) * (t ^ 2) + vo * t + h0

Where,

a: acceleration

vo: initial speed

h0: initial height

By the time the ball hits the ground we have:

h (t) = 0

Substituting values:

0 = - (1/2) * (9.8) * (2 ^ 2) - 25 * 2 + h0

h0 = (1/2) * (9.8) * (2 ^ 2) + 25 * 2

h0 = 69.6 m

Answer:

the building is 69.6 m high