Determining the Time in Air for a Ball Thrown from the Top of a Tower

In the realm of physics and classical mechanics, one fascinating problem is the determination of the time a ball stays in the air when dropped from the top of a tower. If a ball is dropped from a tower of height h and it covers a distance of h/2 in the last second of its motion, our goal is to figure out how long the ball remains in the air. This problem involves the use of equations of motion under constant acceleration due to gravity.

Equations of Motion

The primary equation we will use is the one for distance covered in free fall:

s ut (1/2)gt2

Given the specific conditions of a ball being dropped (initial velocity u 0) and the acceleration due to gravity, g 9.81 m/s2, the equation simplifies to:

h (1/2)gt2

Distance in the Last Second

To understand the problem better, let's focus on the distance covered in the last second of the ball's motion, which is given as h/2. This distance can be calculated using the distance covered in the tth second and the distance covered in the (t-1)th second.

The distance covered in t seconds, st, is given by:

st (1/2)gt2

The distance covered in (t-1) seconds, st-1, is:

st-1 (1/2)g(t-1)2 (1/2)gt2 - 2t (1/2)

The distance covered in the last second is the difference between these two distances:

st - st-1 (1/2)gt2 - [(1/2)gt2 - 2t (1/2)] 2t - (1/2)

Setting this equal to h/2 gives us:

2t - (1/2) h/2

From the equation h (1/2)gt2, we can express g in terms of h and t:

g 2h/t2

Substituting g into the equation for distance in the last second:

(2h/t2)t - (1/2) h/2

Dividing both sides by h (assuming h ≠ 0), we get:

(2t - (1/2))/t2 1/2

Cross-multiplying and simplifying this yields:

4t - (1/2) t2

Expanding and rearranging gives:

t2 - 4t - 2 0

Using the quadratic formula to solve for t:

t (-b ± sqrt(b2 - 4ac))/2a where a 1, b -4, and c -2.