a ball is dropped from a height of 10 ft. if the ball's height after t sec is modeled by the following equation h(t)=-16t^2+10. when does the ball hit the ground?
when it hits the ground, h(t) = 0
0 = -16t^2 + 10
16t^2 = 10
t^2 = 10/16
t = √10/4 or appr .79 seconds
To find out when the ball hits the ground, we need to determine the time at which the height of the ball, represented by the equation h(t) = -16t^2 + 10, becomes zero.
We can set h(t) equal to zero and solve for t:
-16t^2 + 10 = 0
To solve this quadratic equation, we need to use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = -16, b = 0, and c = 10. Substituting these values into the formula, we get:
t = (-0 ± √(0^2 - 4(-16)(10))) / (2(-16))
Simplifying this equation:
t = ± √(0 + 640) / (-32)
t = ± √(640) / (-32)
t = ± √(64 * 10) / (-32)
t = ± 8√10 / (-32)
Simplifying further:
t = ± (√10 / -4)
Since time cannot be negative in this context, we ignore the negative solution. Therefore, the simplified answer is:
t = √10 / -4
So, the ball hits the ground at approximately t = -0.79 seconds, assuming time is being measured since the ball was dropped.