Question

if an object is thrown directly upwards remains in the air for 5.6s before it returns to its original position, how long did it take to reach its maximum height?

Answer 1

okay so an object gets thrown upwards so the half way point of the trip would be the maximum height (before it starts coming back down). if the object stays in the air for a total of 5.6s, then that is the time it takes to go up and come back down. to find the time to maximum height, half the time of the whole trip

5.6s/2 = 2.8s

Answer 2

The study of matter is called physic.

The correct answer is 2.8s.

According to the question, the object is thrown upwards and it falls back to the ground.

The total time taken by the object is 5.6s.

The total distance of the object is shown as follows:-

$H_1 + H_2 = 5.6$

H1 is the vertical upward distance and the H2 is the downward distance.

Both the height is the same then the formula is as follows:-

$2H_1 = 5.6\\H_2 = \frac{5.6}{2}= 2.8$

Hence, the time taken by the object to reach the ground is 2.8sec.

For more information, refer to the link:-

https://brainly.com/question/20293447