Final answer:
To find how many seconds pass before the object hits the ground, we correct the given equation to h = -16t^2 + 900 and solve for t when h = 0, resulting in t = 7.5 seconds.
Step-by-step explanation:
When an object hits the ground, its height (h) is 0. To determine how many seconds pass before the object hits the ground, we need to set the equation h = 16t^2 + 900 equal to 0 and solve for t.
When h = 0, the equation becomes 0 = 16t^2 + 900. Subtracting 900 from both sides gives us 16t^2 = -900. Dividing both sides by 16 results in t^2 = -900/16.
Taking the square root of both sides to solve for t, we remember that we only consider the positive square root since time cannot be negative.
To solve t^2 = -900/16, we simplify the right side: t^2 = -56.25, then we take the square root of both sides, and since the square root of a negative number is not a real number, it seems we have made an error.
Instead, we should expect a positive number on the right side because the distance fallen cannot be negative.
Therefore, we need to correct the equation to h = -16t^2 + 900, as the negative sign indicates the object is falling towards the ground. Now we solve for t.
0 = -16t^2 + 900
16t^2 = 900
t^2 = 900/16
t^2 = 56.25
t = √56.25
t = 7.5 seconds. Therefore, it takes 7.5 seconds for the object to hit the ground.