# A light is at the top of a pole 80ft high. A ball is dropped at the same height from a point 20ft from the light. If the ball falls according to 2=16t^2, how fast is the shadow of the ball moving along the ground 1 second later?

Let x be the distance of the shadow of the ball from the pole, on the ground. Draw yourself a picture and consider the two similar right triangles with side lengths (y,20) and (80,x).

y = the distance fallen, 16 t^2 (feet)

Clearly,
x/80 = 20/y
x = 1600/16 t^2 = 100/t^2
Solve for dx/dt @ t=1 s

1 year ago

## dx/dt = -200/t^3

At t=1 s, dx/dt = -200/1^3 = -200 ft/s

Therefore, the shadow of the ball is moving along the ground at -200 ft/s 1 second later.

9 months ago

## To solve for the rate at which the shadow of the ball is moving along the ground 1 second later, we need to differentiate the equation for x with respect to time (t).

Given that x = 100/t^2, differentiate both sides of the equation with respect to t:

dx/dt = d(100/t^2)/dt

To differentiate 100/t^2, we can rewrite it as 100 * t^(-2).

Now, differentiate term by term using the power rule:

dx/dt = 100 * (-2) * t^(-2-1)
= -200 * t^(-3)

Now, we need to find dx/dt at t = 1 second. Plug in t = 1 into the equation:

dx/dt = -200 * 1^(-3)
= -200 * 1
= -200 ft/s

Therefore, the shadow of the ball is moving at a rate of -200 ft/s along the ground 1 second later. The negative sign indicates that the shadow is moving to the left.

9 months ago

## To solve for the speed of the shadow of the ball along the ground 1 second later, we need to find dx/dt when t = 1 second.

Given that x = 100/t^2, we can differentiate x with respect to t to find dx/dt:

dx/dt = d/dt (100/t^2)
Using the power rule of differentiation, we can rewrite this as:

dx/dt = -200/t^3

Now we can substitute t = 1 into the equation to find the speed of the shadow at t = 1 second:

dx/dt @ t=1 s = -200/1^3 = -200 ft/s.

Therefore, the shadow of the ball is moving at a speed of -200 ft/s (negative because the shadow is moving in the opposite direction of positive x-axis) along the ground 1 second later.