Ball Drop Equation and Time Calculation for Half Height

In physics, particularly kinematics, we often deal with the motion of objects under the influence of gravity. One such problem is determining the time it takes for a ball dropped from a height h to pass through half its original height. This article will guide you through the fundamental steps of solving this classic problem using the equations of motion under gravity.

Introduction to the Problem

When a ball is dropped from a height h and falls freely under gravity, it reaches the ground in a certain time t. Our goal is to find the time t_1 it takes for the ball to reach a height h/2.

Step-by-Step Solution

We will use the standard equations of motion under gravity to solve this problem.

Step 1: Relationship Between Height and Time

The distance s fallen in time t under gravity is given by:

s frac{1}{2} g t^2

where g is the acceleration due to gravity, approximately 9.81 text{m/s}^2.

For the ball dropped from height h, we have:

h frac{1}{2} g t^2

Step 2: Time to Reach Half Height

We need to find the time t_1 when the ball is at height h/2. The distance fallen at this time will be:

s h - frac{h}{2} frac{h}{2}

Using the equation of motion again, we can write:

frac{h}{2} frac{1}{2} g t_1^2

Step 3: Substitute h in Terms of t

From the first equation, we have:

h frac{1}{2} g t^2 text{implies} frac{h}{2} frac{1}{4} g t^2

Substituting this into the second equation:

frac{1}{4} g t^2 frac{1}{2} g t_1^2

Step 4: Simplify the Equation

We can cancel g from both sides assuming g eq 0:

frac{1}{4} t^2 frac{1}{2} t_1^2

Multiplying both sides by 4 to eliminate the fraction:

t^2 2 t_1^2

Step 5: Solve for t_1

Divide both sides by 2:

t_1^2 frac{t^2}{2}

Taking the square root:

t_1 frac{t}{sqrt{2}} approx 0.707t

Conclusion

The time taken to pass through the point at a height h/2 is:

t_1 frac{t}{sqrt{2}} approx 0.707t

Therefore, it takes approximately 0.707t seconds to reach the height h/2.

Note: This solution is based on the standard equations of motion and the acceleration due to gravity. The derivation assumes that air resistance is negligible.

Additional Insights

From basic kinematics, we can also express the relationship as:

S ut frac{1}{2} a t^2

For this specific case where the initial velocity u 0, the acceleration a -g text{(due to gravity, but for simplifying, we use positive g broadly)}:

S -frac{1}{2} g t^2

Therefore, to find the time to reach height h/2, we modify the equation:

-frac{h}{2} -frac{1}{2} g t^2

Solving for t:

t frac{t}{sqrt{2}} frac{t}{sqrt{2}} 0.707t

This confirms our previous solution.

Final Note

Understanding these basic kinematics problems helps in building a strong foundation in physics and can be applied in various real-world scenarios.