Question

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

Answer:

4.9 m

Explanation:

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.

Answer 2

Answer:

v-u/t=g

v-u =2

so that is

2/t=g

T=2SINx/g

T= 2/g

Then =====

2/g=2SINx/g

sinx=1

So height is equal to u*2SIN*2x/2g

sinx=1

So h= from height equation

H=1/2g

Explanation: