Suppose a ball was dropped from a TV tower 2120 feet high. The time it takes an object to free-fall from the top of this tower to the ground can be represented by the equation s = −16t2 + 2120, where s is the distance from the ground in feet and t is the time in seconds. Approximately how long would it take the object to reach the ground? Round your answer to the nearest tenth.
My answer:
If a ball is dropped from 2120 feet, it will take approximately 132.5 seconds to reach the ground.
s equals zero on the ground
0 = −16t2 + 2120
16t2 = 2120 ... t^2 = 132.5 ... this is the SQUARE of the time
take the square root to find the fall time
To find the time it takes for the object to reach the ground, we can set s (distance from the ground) to 0 and solve for t (time).
0 = -16t^2 + 2120
Rearranging the equation:
16t^2 = 2120
Dividing both sides by 16:
t^2 = 132.5
Taking the square root of both sides:
t ≈ √132.5 ≈ 11.5
Therefore, it would take approximately 11.5 seconds for the object to reach the ground.
To find the time it takes for the object to reach the ground, we can set the equation s = -16t^2 + 2120 equal to zero. This is because when the object reaches the ground, its distance from the ground would be zero.
So, we have -16t^2 + 2120 = 0.
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not easy, so we'll use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = -16, b = 0, and c = 2120. Plugging these values into the formula, we get:
t = (-0 ± √(0^2 - 4(-16)(2120))) / (2(-16))
= (± √(0 - (-135040))) / (-32)
= (± √(135040)) / (-32)
= (± 368) / (-32)
Now we can simplify it further by dividing both numerator and denominator by 8:
t = (± 46) / (-4)
= (± 23) / (-2)
Since time cannot be negative in this context, we only consider the positive solution:
t = 23 / 2
= 11.5
Therefore, the object would take approximately 11.5 seconds to reach the ground. Rounding to the nearest tenth gives us 11.5.