Question

During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet) the hammer falls in t sec is s = 16t2. what is the hammer's velocity when it strikes the ground?

Answer 1

T= 24.5 feet per second. That is the velocity it reaches at the end of its fall

Answer 2

Answer:

The hammer's velocity when it strikes the ground is 128 ft/s.

Explanation:

Given that,

The distance (in feet) the hammer falls in t sec is given by the relation as :

$s=16t^2$

The hammer is accidentally dropped from a height of 784 ft. We need to find the hammer's velocity when it strikes the ground. We know that the velocity of an object is equal to :

$v=\dfrac{ds}{dt}\\\\v=\dfrac{d(16t^2)}{dt}\\\\v=32t$

When the hammer strikes ground, s = 256 ft

So,

$16t^2=256\\\\t^2=16\\\\t=4\ s$

So, the velocity of the hammer when it strikes the ground is given by :

$v=32t=32\times 4\\\\v=128\ ft/s$

So, the hammer's velocity when it strikes the ground is 128 ft/s. Hence, this is the required solution.