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:
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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.
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