Question

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

Answer 1

here's your answer _______ ⭐⭐

↪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

$hope \: it \: helps$

Answer 2

For attending maxmium height simple

Very simple = H= u²/2g

H = 20×20/2× 10

H = 20m

as the height was half so maxm. height will be

20×2= 40m