Question

If a body is thrown up with an initial velocity u and covers a maximum height of h, then h is equal to:​

Answer 1

Answer:

Here, initial velocity = u

final velocity = 0 (at max. height, velocity is 0)

height covered = h

acceleration due to gravity = g

Using the law of kinematics : v² - u² = 2as

we get -u² = -2gh (considering upward direction as positive, "g" is always downward)

Thus, we get h = u²/2g

PS: If you find the answer helpful, please mark it as brainliest

Answer 2

Explanation:

initial velocity=u

final velocity =0(at maximum height it is 0)

Height=h metres

a=g

By applying newton's third law of motion

V^2=U^2+2gh

2gh=V^2-U^2

By taking g=9.8 m/s^2

2×9.8×h=V^2-U^2

h=V^2-U^2/19.8

Hopefully this will help you

If this is correct then it's up to you