When an object is projected upwards with an initial velocity u, it follows a parabolic trajectory. Let's consider the motion of this object and determine the time it takes for the object to pass through a certain point during its return journey.



When the object is projected upwards, it experiences a constant acceleration due to gravity in the opposite direction of its motion. This acceleration is denoted by g and is equal to approximately 9.8 m/s².

During its ascent, the object slows down due to the opposing force of gravity until it reaches its highest point. At this point, the object momentarily comes to rest before starting its descent.



To determine the time it takes for the object to reach its highest point, we can use the equation of motion:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

At the highest point, the final velocity v becomes zero. Therefore, the equation becomes:

0 = u + gt

Solving for t, we find:

t = -u/g

Since time cannot be negative, we take the magnitude of t:

t = u/g

This is the time it takes for the object to reach its highest point.



During the descent, the object follows the same path it took during the ascent, but in the opposite direction. Therefore, the time it takes for the object to pass through the same point during the return journey is equal to twice the time taken to reach the highest point.

t₁ = 2 * (u/g)

Thus, the time after which the body passes through the same point during the return journey is equal to twice the time taken to reach the highest point.



When an object is projected upwards with an initial velocity u, the time it takes for the object to pass through a certain point during its return journey is twice the time taken to reach the highest point. This can be calculated using the equation t₁ = 2 * (u/g).

T1+ t2= 2u/g so....t2= 2u/g-t1