Question

Airtel is thrown vertically returns to the thrower after 6 second find the velocity with which it was throw up take maximum height it reaches (g= 9.8)

Answer 1

Let the initial velocity with which it is thrown be u.

Total time taken = 6 seconds

Time taken for the half journey =  3 seconds

The velocity at the highest point (v) = 0 m/s

Acceleration due to gravity = - 9.8 m/s^2

Now, using first equation of motion:

v = u + at

0 = u - 9.8 × 3

u = 29.4 m/s

Hence, it is thrown with a velocity of 29.4 m/s

Answer 2

_/\_Hello mate__here is your answer--

_______________________

The ball takes a total of 6 s for its upward and downward journey. (Time of ascent is equal to the time of descent.)

Hence, it has taken 3 s to reach at the maximum height.

v = 0 m/s

g = −9.8 ms−2

Using equation of motion,

v = u + at

⇒0 = u + (−9.8 × 3)

⇒ u = 9.8 × 3 = 29.4 m/s

Hence, the ball was thrown upwards with a velocity of 29.4 m/s.

I hope, this will help you.☺

Thank you______❤

_________________________❤