# Question Video: Studying the Motion of an Object Projected Upwards That Falls on a Building Mathematics

A body was projected upwards at 34.3 m/s from the ground. It fell on a roof of a building 4.5 seconds after it was projected. Find, to the nearest two decimal places, the height of the building āā and the maximum height the body reached āā. Take the acceleration due to gravity š = 9.8 m/sĀ².

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### Video Transcript

A body was projected upwards at 34.3 meters per second from the ground. It fell on a roof of a building 4.5 seconds after it was projected. Find, to the nearest two decimal places, the height of the building ā one and the maximum height the body reached ā two. Take the acceleration due to gravity š equal to 9.8 meters per square second.

We are told that the body is projected upwards with initial velocity 34.3 meters per second. It landed on the roof of a building after 4.5 seconds. The acceleration due to gravity is 9.8 meters per square second. This is acting downwards. In order to solve this problem, we will use our equations of motion or SUVAT equations. The first part of the question wants us to calculate the height of the building ā one. This will be the displacement of the body when it lands on the roof. Therefore, š  is equal to ā one. Our initial velocity š¢ is 34.3. š is equal to negative 9.8 as the body was projected upwards against gravity. š” is equal to 4.5.

We can use the equation š  is equal to š¢š” plus a half šš” squared to calculate the value of ā one. Substituting in our values gives us ā one is equal to 34.3 multiplied by 4.5 plus a half multiplied by negative 9.8 multiplied by 4.5 squared. This simplifies to 154.35 minus 99.225. ā one is therefore equal to 55.125. To two decimal places, the height of the building is 55.13 meters.

The second part of this question asks us to calculate the maximum height of the body ā two. At the maximum height, we know that the velocity is equal to zero. This time our displacement š  is equal to ā two. š¢ is still equal to 34.3, š negative 9.8. And as mentioned, š£ is equal to zero. We will use the equation š£ squared is equal to š¢ squared plus two šš  to calculate ā two. As zero squared is equal to zero, this is equal to 34.3 squared plus two multiplied by negative 9.8 multiplied by ā two. 34.3 squared is equal to 1176.49.

Rearranging this equation, we see that this is equal to 19.6ā two. Dividing both sides of this equation by 19.6 gives us ā two is equal to 60.025. Once again, we need to round this to two decimal places, giving us a value of ā two of 60.03 meters. The height of the building is 55.13 meters. And the maximum height the body reached is 60.03 meters.