a ball is dropped from the top of a building that is known to be 256 feet high.the formula for finding the height of the ball at any time t is h=256-16t^2,where t is measured in seconds.how many seconds will it take for the ball to hit the ground?
it hits the ground when the height is zero. So, just solve
256-16t^2 = 0
(16-4t)(16+4t) = 0
...
To find out how many seconds it will take for the ball to hit the ground, we need to set the height equation h = 256 - 16t^2 equal to 0, since the ball will hit the ground when the height is 0.
So, we have the equation: 256 - 16t^2 = 0
To solve for t, we can divide both sides of the equation by -16 to isolate t^2:
-16t^2 = -256
Dividing both sides by -16:
t^2 = (-256) / (-16)
Simplifying further:
t^2 = 16
Taking the square root of both sides:
t = ±√16
Since time cannot be negative in this context, we take the positive square root:
t = √16
Therefore, the ball will hit the ground after 4 seconds.
To find out how many seconds it will take for the ball to hit the ground, we need to determine when the height of the ball is zero. In other words, we need to solve the equation h = 256 - 16t^2 for t when h = 0.
The equation is h = 256 - 16t^2. We set h to zero:
0 = 256 - 16t^2
To solve for t, we rearrange the equation:
16t^2 = 256
Divide both sides of the equation by 16:
t^2 = 16
To isolate t, we take the square root of both sides:
t = √16
Since 16 has two possible square roots, positive 4 and negative 4, we discard the negative value. Thus, t = 4.
Therefore, it will take 4 seconds for the ball to hit the ground.