Susan throws a softball upward into the air at a speed of 32 feet per second from a 40-foot platform. The height of the ball after seconds is given by the function h(t)= -16t² + 32t + 40
When does the softball hit the ground?
To find when the softball hits the ground, we need to find the time when the height of the ball is zero (ground level). So, we set h(t) = 0 and solve for t:
-16t² + 32t + 40 = 0
Divide by -16 to simplify:
t² - 2t - 2.5 = 0
Now, we need to solve this quadratic equation. We can either factor it or use the quadratic formula. Using the quadratic formula:
t = (-(-2) ± √((-2)² - 4(1)(-2.5))) / 2(1)
t = (2 ± √(4 + 10)) / 2
t = (2 ± √14) / 2
t = (2 ± √14) / 2
t = 1 ± √14/2
The two possible solutions are t = 1 + √14/2 and t = 1 - √14/2. However, the negative solution does not make sense in this context (time cannot be negative), so we will consider t = 1 + √14/2 as the time when the softball hits the ground.
Therefore, the softball hits the ground at approximately 2.54 seconds.