a ball dropped from the top of a building has a height of s = 256 - 16t^2 meters after t seconds. How long does it take the ball to reach the ground? What is the ball's velocity at the moment of impact?

Answer 1

To find out how long it takes for the ball to reach the ground, we need to find the time when the height, s, is equal to zero. The equation for s is given as s = 256 - 16t^2.

Setting s equal to zero, we can solve for t:

0 = 256 - 16t^2

Rearranging the equation:

16t^2 = 256

Dividing both sides by 16:

t^2 = 16

Taking the square root of both sides:

t = ±4

Since time cannot be negative in this context, we take the positive value:

t = 4

So, it takes the ball 4 seconds to reach the ground.

To find the ball's velocity at the moment of impact, we need to find the derivative of the position equation.

The derivative of s is given by:

s' = d/dt(256 - 16t^2)

s' = -32t

Evaluating s' at t = 4:

s'(4) = -32(4) = -128

Therefore, the ball's velocity at the moment of impact is -128 meters per second.