Question

If magnitude of change in velocity of particle
projected from ground till it lands on ground as
shown in figure is 2m/s, find maximum height of the
projectile. [g=acceleration due to gravity]
(1) m/g
(2) m/4g
(3)2m/g
(4)m/2g​

Answer 1

Answer:

Given :

An object projected from ground in upward direction.

When the objects reaches ground from the maximum height it can attain, the velocity is 2 m/s.

To Find :

Maximum height attained in the projection.

Solution :

Considering downward direction as direction of motion,

When the object attain its maximum height it

can, the velocity is 0.

So, u = 0 m/s

As, given in question, when the object reaches the ground, the velocity is 2 m/s.

So, v = 2 m/s

Acceleration will be considered positive here (As the acceleration is in the direction of motion).

So, g = 9.8 m/s²

Considering maximum height to be found as

X.

So, h = x

By third Kinematical equation of motion (Vertical motion) :

v²-u² = 2*g*h

2²-0² = 2*9.8*x

4-0 = 19.6*x

4/19.6 = x

4.9 = X

Hence, the maximum height attained by the object in the projection is 4.9 m

If you feel it helpful - Mark me as BRAINLIEST