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.
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?
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