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?
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.