Calculating the Time for a Ball Thrown Upward to Hit the Ground from a Height

The problem of how long it takes for a ball to fall from a height to the ground is a common inquiry in the field of physics. In this article, we will break down the process of solving this problem using several methods and equations, including kinematic equations and the quadratic formula. Specifically, we will examine the scenario where a ball is thrown upward from a building, and determine how long it takes for the ball to hit the ground.

Understanding the Initial Conditions and Equations

Consider a ball thrown upward from a 35 meter tall building with an initial velocity of 22 meters per second. We are interested in determining the time it takes for the ball to hit the ground. Let's start by calculating the time taken by the ball to reach its maximum height, and then the time taken to fall back to the ground from that height.

First, we can use the equation for velocity under constant acceleration (kinetic motion):

v u at

When the ball reaches its maximum height, its velocity becomes 0:

0 22 - 9.81t

Solving for t:

t 22 / 9.81 ≈ 2.25 seconds

Next, we need to verify the distance the ball travels upward.

v^2 - u^2 2as

Substituting the known values:

0 - 22^2 2(-9.81)s

Calculating the distance:

s 22^2 / (2 * 9.81) 24.25 meters

The total distance the ball travels upward from the ground is:

35 24.25 59.25 meters

Solving for Total Time Using the Standard Method

We can now use the kinematic equation to find the total time it takes for the ball to hit the ground:

h h_0 v_0t - (1/2)gt^2

Substituting the known values into the equation:

0 35 22t - (1/2)(9.81)t^2

Rewriting for clarity:

4.905t^2 - 22t - 35 0

Using the quadratic formula to solve for t (where a 4.905, b -22, and c -35):

t (-b plusmn; sqrt(b^2 - 4ac)) / (2a)

Calculating the discriminant:

b^2 - 4ac (-22)^2 - 4 * 4.905 * (-35) 484 686.7 1170.7

Substituting back into the quadratic formula:

t (22 plusmn; sqrt(1170.7)) / (2 * 4.905)

Calculating the two possible values for t:

t ≈ (22 34.24) / 9.81 ≈ 5.73 seconds (taking the positive value)

Taking the positive value, we find that the total time it takes for the ball to hit the ground is approximately 5.73 seconds.

Alternative Method for Solution

Alternatively, we can use a simpler approach to find the time the ball takes to fall freely from the height calculated above. Given the initial velocity and time to reach the maximum height, we can determine the height and then the time it takes to fall back to the ground:

h ut - (1/2)gt^2

For the height above the building:

h 15(15/g) - (1/2)g(15/g)^2

Which simplifies to:

h 225/g - 225/(2g) 225/(2g) meters

The ball falls freely from a height of:

30 225/(2g) 600225/(100) 825/100 8.25 meters

Using the equation for freely falling time:

h (1/2)gt^2

Substituting the values:

8.25 (1/2)(9.81)t^2

Calculating the time:

t^2 (2 * 8.25) / 9.81 ≈ 1.67 seconds

Total time taken:

2.87 1.5 4.37 seconds (approximately)

Therefore, we can conclude that using different methods, we arrive at approximately 4.37 to 5.73 seconds.