Answer: The ball reaches a height of 68 feet after 1.7 seconds.
Explanation: To solve this problem, we can use the following formula:
h(t) = -16t^2 + vt + h
where h(t) is the height of the ball at time t, v is the initial velocity (in this case, 80 ft./sec.), and h is the initial height (in this case, 4 feet).
We want to find out when the height of the ball is 68 feet, so we can set h(t) = 68 and solve for t:
68 = -16t^2 + 80t + 4
Rearranging and simplifying:
16t^2 - 80t - 64 = 0
Dividing both sides by 16:
t^2 - 5t - 4 = 0
Factoring:
(t - 4)(t + 1) = 0
So t = 4 or t = -1. We can discard the negative value, since time cannot be negative. Therefore, the ball reaches a height of 68 feet after t = 4 seconds - the time at which it reaches maximum height.
However, we also know that the initial height of the ball is 4 feet, so we need to subtract this from the maximum height to get the actual height above the ground:
max height = h(4) = -16(4)^2 + 80(4) + 4 = 132 feet
So the ball first reaches a height of 68 feet on its way down, after reaching a maximum height of 132 feet. We can use the formula again to find when the ball reaches 68 feet on its way down:
68 = -16t^2 + 80t + 132
Rearranging and simplifying:
16t^2 - 80t + 64 = 0
Dividing both sides by 16:
t^2 - 5t + 4 = 0
Factoring:
(t - 1)(t - 4) = 0
So t = 1 or t = 4. We discard the value t = 4, since that is when the ball reaches maximum height. Therefore, the ball reaches a height of 68 feet again, on its way down, after t = 1 second.
Therefore, the ball reaches a height of 68 feet after 1.7 seconds (1 second on the way down + 0.7 seconds to reach maximum height on the way up).
Hope this helps, and have a great day!