Question Video: Finding the Distance Covered during the Last Second of a Body Free Falling Mathematics

Video Transcript

A body was projected vertically downwards at 66.9 meters per second from the top of a tower that is 457 meters high. Determine the distance that the body covered in the last second before it struck the ground. Take the acceleration due to gravity 𝑔 equal to 9.8 meters per square second.

We are told that the initial velocity of the body is 66.9 meters per second. The tower is 457 meters high. The acceleration is 9.8 meters per square second. We will let the velocity at which the body hits the ground be 𝑣 meters per second. We can answer this question using the equations of motion or SUVAT equations. In the entire journey of the body, we know that the displacement 𝑠 is 457 meters. This is the height of the tower. 𝑢 is equal to 66.9 and 𝑎 is equal to 9.8.

We can calculate the value of 𝑣 using the formula 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. Substituting in our values gives us 𝑣 squared is equal to 66.9 squared plus two multiplied by 9.8 multiplied by 457. Typing the right-hand side into our calculator gives us 13432.81. Square rooting both sides of this equation gives us a value of 𝑣 of 115.9. The velocity of the body as it hits the ground is 115.9 meters per second.

We need to calculate the distance covered in the last second before the body struck the ground. As this is a one-second period, 𝑡 is equal to one. The final velocity 𝑣 is 115.9. 𝑎 is still equal to 9.8. If we let the distance equal 𝑥, we can calculate this using the formula 𝑠 is equal to 𝑣𝑡 minus a half 𝑎𝑡 squared. Substituting in our values gives us 𝑥 is equal to 115.9 multiplied by one minus a half multiplied by 9.8 multiplied by one squared. This simplifies to 115.9 minus 4.9. 𝑥 is therefore equal to 111.

The distance covered in the last second before the body struck the ground is 111 meters.