Question

An object is thrown downward from the top of a 220 ft. building with an initial velocity of 26 ft/sec. Write the position equation for the movement described. What is the velocity at 1 second?

Answer 1

The position equation is y = 220 + 26t + ½(-32)t^2, and the velocity at 1 second is -6 ft/s, pointing downward.

Step-by-step explanation:

To solve for the position equation of an object thrown downward with an initial velocity, we use the kinematic equation for uniformly accelerated motion:

y = y0 + v0t + ½at2

Where y is the final position, y0 is the initial position (the height of the building), v0 is the initial velocity, t is the time, and a is the acceleration due to gravity (32 ft/s2 because we are using feet and seconds).

Substituting the given values:

y = 220 + 26t + ½(-32)t2

This is the position equation.

To find the velocity at 1 second, we use the velocity equation:

v = v0 + at

Substituting v0 = 26 ft/s and a = -32 ft/s2:

v = 26 + (-32)(1) = -6 ft/s

The negative sign indicates that the velocity is directed downward.