A particle is projected upwards with velocity 20 m/s. Simultaneously another particle is projected with velocity…..

Ans: (a)

Sol: Time taken by First particle to reach maximum height is

t = 20/10 = 2 sec

Hence after 2 sec first particle is at maximum height .

Max height

$ \displaystyle h_1 = \frac{u^2}{2 g} $

$ \displaystyle h_1 = \frac{20^2}{2 \times 10} = 20 m $

Time of flight of 2nd particle is

$ \displaystyle t’ = \frac{2 \times 20 \sqrt{2} sin45 }{10} = 4 sec $

Hence , After 2 sec it is at max. height

Max height reached is

$ \displaystyle h_2 = \frac{(20\sqrt{2})^2 sin^2 45}{2\times 10}= 20 m $

Hence , Both the particle is at same height .

Now , horizontal distance covered in 2 sec (i.e. particle is at max height) = R/2

$ \displaystyle \frac{R}{2} = \frac{1}{2}\times \frac{(20\sqrt{2})^2 sin90}{10} $

= 40 m