Question

A particle is projected vertically upwards. The time
corresponding to a location at height h while ascending
and while descending are 3 s and 5 s respectively, then
velocity of projection is (g = 10 m/s2)

Answer 1

Given:

Particle is projected upwards such that it crossed same height h at 3 seconds (while ascending) and again at 5 seconds (while descending)

To find:

Velocity of projection of particle.

Concept:

We know that equal distance is travelled by any object at a particular second during ascending and descending.

For example : Distance travelled in the last second of ascent will be equal to the distance travelled in 1st second of descent.

Calculation:

We can divide the linear trajectory into parts travelled in a particular second as shown in the attached diagram :

So , we can see that the particle crosses height h at 3 seconds and at 5 seconds.

Then , we can say that the particle reached maximum height at 4 seconds.

Let initial velocity be u ;

Applying First Equation of Kinematics :

$\therefore \: v= u + at$

$= > 0 = u + \{( - 10) \times 4 \}$

$= > u = 40 \: m {s}^{ - 1}$

So the object has been thrown upwards with 40 m/s .