Antiderivative Application

Question

A rock is thrown downward with a velocity 9ft/sec from a bridge 111ft above the water. How fast is the rock moving when it hits water. Round to three decimal places

Answer 1

Gravity is 32.17 ft/sec^2

We'll use the following to determine our position

y(t) is the equation to determine the position, y, based on t, time.

g = acceleration of gravity, or 32.17

v₀ = initial velocity

y₀ = initial position

We can determine the placement of the rock with the following equation:

y(t) = (-1/2)g*t² - v₀*t + y₀

y(t) = (-1/2)32.17t^2 - 9t + 111

Some notes:

* I chose y because we're moving in a vertical direction.

* 32.17t^2 because gravity is pulling down at 32.17 feet per second per second.

* -9t because rock thrown downwards at constant 9 feet per second.

* 111 because it starts 111 feet above.

Let's find out when the rock hits the river. This happens when at t when y(t) = 0.

0 = -16.09t^2 - 9t + 111

The fastest way to do this is with a calculator, physical or online. Graph the function and use it to find the zero(s). The roots are t≈-2.921 and t≈2.362

We are looking for answers in real time, so the relevant answer is t≈2.362

Now let's find out what the velocity of the rock is at the time it hits the water.

Derive the position function to find the velocity function:

s'(t) = v(t) = -32.17t - 9

Now input the time that we found when the rock hits the water into the velocity equation:

v(2.362) = -32.17(2.362) - 9

= -75.99 - 9

= -84.99 ft/sec (or 84.99 ft/sec downwards)

Also, no antiderivatives needed as far as I can tell.