### Video Transcript

A body was projected vertically upwards from the ground, and it took 157 seconds to return to the ground. Find the time π‘ one for which the body was ascending and time π‘ 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 π£ is equal to π’ plus ππ‘. π£ squared is equal to π’ squared plus two ππ . π is equal to π’ plus π£ divided by two multiplied by π‘. π is equal to π’π‘ plus a half ππ‘ squared. And π is equal to π£π‘ minus a half ππ‘ squared. In all of these equations, π’ is equal to the initial velocity, π£ is the final velocity, π is the acceleration, π‘ is the time, and π is the displacement.

This question, however, doesnβt 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 π‘ one is equal to π‘ two. As the total time was 157 seconds, π‘ one plus π‘ two is equal to 157. Replacing π‘ two with π‘ one in this equation gives us two π‘ one is equal to 157. Dividing both sides of this equation by two gives us π‘ one is equal to 78.5. As π‘ one is equal to π‘ 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.