David drops a soccer ball off a building. The building is 75 meters tall. (a) What is the height of the ball to the nearest tenth of a meter exactly 1 seconds after he drops the ball?

Answer 1

Define your coordinate system. In this case, I'll let the positive y-axis point upward. At t = 0, the ball is at y = 75. y = 0 is the ground. We need to find the y-coordinate after 1 s and that will be our height.

We know that the acceleration due to gravity of the earth is g.

Let a(t) be the acceleration of the ball at any time t.

a(t) = -g (The negative means that the ball is accelerating towards negative infinity on the y-axis)

Integrate acceleration to find velocity.

v(t) = -gt + C

The initial velocity allows us to clear C (it's 0).

v(t) = -gt

Integrate velocity to find position as a function of time.

r(t) = -0.5gt² + C

Use the initial position to clear C.

r(0) = -0.5g(0)² + C = 75

C = 75

So, r(t) = -0.5gt² + 75

Find the position after 1 s.

r(1) = -0.5g(1)² + 75 = 70.1

The ball is at a height of 70.1 m after he drops the ball and before it is eaten by the dragon.