Use the position function s(t) = −16t² + 400, which gives the height (in feet) of an object that has fallen for t seconds from a height of 400 feet. the velocity at time t = a seconds is given by . a construction worker drops a full paint can from a height of 500 feet. when will the paint can hit the ground? at what velocity will the paint can impact the ground?

Using movement concepts, it is found that:

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The height of the object after t seconds, dropped from a height of 500 feet, is given by:

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It hits the ground when , thus:

The object hits the ground after 5.59 seconds.

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The velocity is the derivative of the position, thus:

The velocity when it impacts the ground is v(5.59), thus:

The object hits the ground at a velocity of -178.88 feet per second.

A similar problem is given at link

160m/s

Step-by-step explanation:

The object can hit the ground when t = a; meaning that s(a) = s(t) = 0

So, 0 = -16a² + 400

16a² = 400

a² = 25

a = √25

a = 5 (positive 5 only because that's the only physical solution)

The instantaneous velocity is

v(a) = lim(t->a) [s(t) - s(a)]/[t-a)

Where s(t) = -16t² + 400

and s(a) = -16a² + 400

v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)

v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)

v(a) = lim (t->a) -16(t² - a²)(t-a)

v(a) = -16lim t->a (t²-a²)(t-a)

v(a) = -16lim t->a (t-a)(t+a)/(t-a)

v(a) = -16lim t->a (t+a)

But a = t

So, we have

v(a) = -16lim t->a 2a

v(a) = -32lim t->a (a)

v(a) = -32 * 5

v(a) = -160

Velocity = 160m/s