Question

A ball is projected vertically upward. Its speed at half of maximum height is 20m/s.the maximum height attained by it is

Answer 1

acceleration due to gravity=-10m/s
speed or v at half distance=20m/s
height=?
v^-u^=2as
20^-0^=2*10m/s*s
400=20m/s*s
s=20m at half distance
total distance =20+20=40m

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Answer 2

Use equation of motion v^2 - u^2 = 2aS

So, half of the upwards journey
H/2 = (v^2 - u^2) / (-2g)
H/2 = (20^2 - u^2) / (-2g)

Maximum height is
H = u^2 / (2g)

Substitute value of H in first equation

u^2 / (4g) = (20^2 - u^2) / (-2g)
u^2 / 2 = -(400 - u^2)
u^2 / 2 = 400
u^2 = 800

Substitute u^2 = 800 in 2nd equation

H = u^2 / (2g)
= 800 / (2 × 10)
= 40 m

Maximum height attained by it is 40 m