Question

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?

Answer 1

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