Question

During the construction of an office building, a hammer is accidentally dropped from a height of 256 ft. The distance (in feet) the hammer falls in t sec is given by s = 16t². What is the hammer’s velocity when it strikes the ground? What is the hammer's acceleration?

Answer 1

The hammer's velocity when it strikes the ground is 32 feet per second, and its acceleration is 32 feet per second squared.

Step-by-step explanation:

To determine the velocity and acceleration of the falling hammer, we can use the kinematic equation that relates displacement, initial velocity, time, and acceleration:
`\(s = ut + (1)/(2)at^2\)`, where (s) is the distance fallen, (u) is the initial velocity, (t) is time, and (a) is acceleration.

Given the equation
`\(s = 16t^2\)` for the distance fallen, we can find the velocity by taking the first derivative of (s) with respect to (t), giving us the velocity function
`\(v = (ds)/(dt) = 32t\)`. To find the velocity when the hammer strikes the ground
`(\(s = 256\))`, substitute (s = 256) into the equation and solve for (t):
`\(256 = 16t^2\)`. Solving for (t), we get (t = 4). Substituting (t = 4) into the velocity function, we find
`\(v = 32 * 4 = 128\)` feet per second.

The acceleration is the derivative of velocity with respect to time,
`\(a = (dv)/(dt) = 32\)`, which is constant. Thus, the hammer's acceleration is (32) feet per second squared, and the velocity when it strikes the ground is (128) feet per second.