A ball is dropped from the same height, but 20
feet away from the side of the building. The height y of the
ball at time t is given by y(t) = 64 − 16t2. How fast is the
shadow of the ball moving along the ground after 1 second?
The velocity of the ball after time t is given by:
y'(t)=-32t
The shadow cannot be traced unless the position of the sun or light source is known.
To find out how fast the shadow of the ball is moving along the ground after 1 second, we need to determine the rate of change of the distance between the ball and the point where its shadow is cast on the ground.
Let's start by sketching the situation. We have a ball dropped from a height with a given position function y(t) = 64 - 16t^2, where t represents time. The shadow of the ball on the ground forms a right triangle with the ball and the building.
Let's assume that the length of the shadow is x(t), and we want to find dx/dt, the rate of change of x with respect to time.
Since the ball is dropped 20 feet away from the side of the building, the horizontal distance between the ball and the base of the building is a constant value, let's call it d (d = 20 feet in this case).
In the right triangle formed by the ball, its shadow, and the ground, we can relate the position functions of the ball and its shadow using the similar triangles property:
y(t) / x(t) = (height of the building) / (distance between the ball and the base of the building)
The height of the building is not given, but it cancels out when we differentiate both sides of the equation with respect to time t:
(dy/dt) / x(t) - y(t) * (dx/dt) / x(t)^2 = 0
Now, we can solve this equation to find dx/dt, the rate at which the shadow is moving along the ground after 1 second.
First, let's find dy/dt, the rate at which the ball is falling after 1 second. Differentiating y(t) = 64 - 16t^2 with respect to t:
dy/dt = -32t
Next, substitute the values for y(t), dy/dt, and t into the equation:
-32 / x(t) - (64 - 16t^2) * (dx/dt) / x(t)^2 = 0
At t = 1 second, we have:
-32 / x(1) - (64 - 16(1)^2) * (dx/dt) / x(1)^2 = 0
Simplify the equation by substituting the value of x(1) = d = 20 feet and solving for dx/dt:
-32 / 20 - (64 - 16) * (dx/dt) / 20^2 = 0
Simplifying further:
-32 / 20 - 48 * (dx/dt) / 400 = 0
Now, solve for dx/dt:
-32 / 20 = 48 * (dx/dt) / 400
Solving for dx/dt:
(dx/dt) = (-32 / 20) * (400 / 48)
(dx/dt) = -16/3
Therefore, the shadow of the ball is moving along the ground at a rate of -16/3 feet per second after 1 second.