Post a New Question


posted by on .

A ball is thrown upwards, from a building. It's height(h), in meters above the ground seconds(t) after being thrown is given by h(t)=-5t^2+15t+50

a) How tall is the building? (explain your answer)

b) How long does it take the ball to reach the ground?

c) For how long is the ball 55m above the ground?

  • Physics - ,

    You know that the initial height is the constant term. So, the building is 50m tall

    The ball hits the ground when the height is zero. So, just solve for t when h=0

    I assume you want to know how long the ball is at least 55m above the ground. It is only at 50m for an instant. So, we want

    -5t^2+15t+50 >= 55
    1/2 (3-√5) <= t <= 1/2 (3+√5)^2%2B15t%2B50+%3E%3D+55

  • Physics - ,

    a) h(t)=-5t^2+15t+50

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question