Question

A body thrown up with initial velocity u covers a maximum height of h, then the maximum height attained by the body if it is thrown up with initial velocity u/2 is:

Answer 1

Answer:

h/4.

Step-by-step explanation:

first case

initial velocity is u.

height is h.

final velocity is 0.

acceleration due to gravity is -g.

we know v^2 - u^2 = 2as

0^2 - u^2 = 2(-g)h

u^2 = 2gh. this is equation 1

second case

initial velocity is u/2

height is 'y'

final velocity is 0.

acceleration due to gravity is -g.

we know v^2 - u^2 = 2as

0^ - (u/2)^2 = 2(-g)y

(u^2/4) = 2gy

y = u^2/8g

substitute equation 1 in the above equation.

y = 2gh/8g

y = h/4.