Savagehailey
contestada

A student drops a rock over the edge of the well and hears it splash into water after 3 seconds. Write a function in the form h(t) = − 16t2 + v0 t + h0 to determine the height of the rock above the bottom of the well t seconds after the student drops the rock. What is the distance from the surface of the water to the bottom of the well?

A student drops a rock over the edge of the well and hears it splash into water after 3 seconds Write a function in the form ht 16t2 v0 t h0 to determine the he class=

Respuesta :

Answer:

h(t) = -16t² + 200

distance = 56 ft

Step-by-step explanation:

First of all, we know that the initial velocity (v0) of the rock is 0 ft/s because the rock was dropped. We also know that the initial height of the rock is 200 ft because it was dropped at the top of the well which is 200 ft high. Plugging this information into our quadratic function we get h(t) = -16t² + 200.

Now we need to find the distance from the surface of the water to the bottom of the well. This distance is equal to the height of the rock at time equals 3 seconds. Therefore, all we need to do is find h(3) = -16(3)² + 200 = 56 ft.