Calculating the Height of a Building Using Kinematic Equations for a Ball Thrown Upwards

Understanding and applying kinematic equations can help us solve problems involving motion under uniform acceleration, such as those involving a ball thrown upwards from the top of a building. In this article, we will explain the process in detail, using the example of a ball thrown upwards with an initial velocity of 19.6 m/s and returning to the ground after 5 seconds.

Objectives

Learn how to use kinematic equations to find the height of a building. Understand the significance of initial velocity and acceleration due to gravity. Practice solving real-world problems using these equations.

Problem Description

A ball is thrown upwards from the top of a building with an initial velocity of 19.6 m/s. The ball reaches the ground after 5 seconds. We need to determine the height of the building.

Relevant Equations

The kinematic equation we need is:

[ s ut frac{1}{2} a t^2 ]

Where:

( s ) is the displacement height of the building (which is negative because it moves downwards). ( u ) is the initial velocity (19.6 m/s upwards). ( a ) is the acceleration due to gravity (-9.81 m/s2, negative because it acts downwards). ( t ) is the time (5 seconds).

Step-by-Step Calculation

Let's now substitute the values into the equation:

( u 19.6 ) m/s ( a -9.81 ) m/s2 ( t 5 ) s

The equation now looks like this:

[ s 19.6 cdot 5 frac{1}{2} cdot (-9.81) cdot 5^2 ]

Breaking it down further:

( u cdot t 19.6 cdot 5 98 ) m ( frac{1}{2} a t^2 frac{1}{2} cdot (-9.81) cdot 25 -122.625 ) m

Combining these terms gives:

[ s 98 - 122.625 -24.625 ) m

The negative sign indicates a downward displacement from the point of release.

Height of the Building

Thus, the height of the building is approximately ( 24.63 ) meters.

Additional Examples and Solutions

For verification and further understanding, let's look at another example:

Example 1

Given: ( u 10 ) m/s, ( a -10 ) m/s2, ( t 5 ) s.

The equation ( s ut frac{1}{2} a t^2 ) still applies, but since the ball is not dropped freely and has an initial velocity, we need to use the complete equation:

[ s 10 cdot 5 frac{1}{2} (-10) cdot 5^2 ]

Calculate each term:

( u cdot t 10 cdot 5 50 ) m ( frac{1}{2} a t^2 frac{1}{2} cdot (-10) cdot 25 -125 ) m

Combining these terms:

[ s 50 - 125 -75 ) m

The height of the tower is ( 75 ) meters.

Example 2

Given: ( u 10 ) m/s, ( a -9.81 ) m/s2, ( t 5 ) s.

We will split the journey into two parts: the upward and downward motion.

Upward distance: ( v^2 - u^2 2as ) ( 0 - 10^2 2 cdot (-9.81) cdot s ) ( s frac{100}{19.62} 5.1 ) m Time taken upwardly: ( v u at ) ( 0 10 - 9.81t ) ( t frac{10}{9.81} 1.02 ) s Time taken downward: ( 5 - 1.02 3.98 ) s

Downward distance: ( s ut frac{1}{2} at^2 )

( s 0 frac{1}{2} cdot 9.81 cdot (3.98)^2 77.7 ) m

Total height of the tower: ( 77.7 - 5.1 72.6 ) m

Conclusion

In conclusion, we have seen how to apply kinematic equations to solve a variety of problems involving motion, including the height of a building. Understanding these equations and their applications can be very useful in real-world scenarios.