Question Video: Studying the Motion of an Upward Projected Body until It Comes Back to the Same Point Mathematics • Second Year of Secondary School

Video Transcript

A body was projected vertically upwards from the ground, and it took 157 seconds to return to the ground. Find the time 𝑡 one for which the body was ascending and time 𝑡 two for which it was descending.

Our initial thoughts when dealing with a problem like this might be to use our equations of motion or SUVAT equations. These are 𝑣 is equal to 𝑢 plus 𝑎𝑡. 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. 𝑠 is equal to 𝑢 plus 𝑣 divided by two multiplied by 𝑡. 𝑠 is equal to 𝑢𝑡 plus a half 𝑎𝑡 squared. And 𝑠 is equal to 𝑣𝑡 minus a half 𝑎𝑡 squared. In all of these equations, 𝑢 is equal to the initial velocity, 𝑣 is the final velocity, 𝑎 is the acceleration, 𝑡 is the time, and 𝑠 is the displacement.

This question, however, doesn’t require any of these formulas. This is because when we model a body as a particle and ignore air resistance, the time that it is ascending will be equal to the time it is descending. This means that 𝑡 one is equal to 𝑡 two. As the total time was 157 seconds, 𝑡 one plus 𝑡 two is equal to 157. Replacing 𝑡 two with 𝑡 one in this equation gives us two 𝑡 one is equal to 157. Dividing both sides of this equation by two gives us 𝑡 one is equal to 78.5. As 𝑡 one is equal to 𝑡 two, we can conclude that they are both equal to 78.5. The time for which the body is ascending and the time for which it was descending are both 78.5 seconds.