A particle is projected at an angle `60^@` with horizontal with a speed `v = 20 m//s`. Taking `g = 10m//s^2`. Find the time after which the speed of the particle remains half of its initial speed.

A particle is projected at an angle `60^@` with horizontal with a spee

1.	A particle is projected at an angle `60^@` with horizontal with a speed `v = 20 m//s`. Taking `g = 10m//s^2`. Find the time after which the speed of the particle remains half of its initial speed.
Answer» Correct Answer - C `v_x = u_x = 20 cos 60^@ = 10 m//s` Given, `v=u/2` `:. 4 v^2 = u^2` or `4(v_(x)^2 + v_(y)^2) = u^2` `:. 4[(10)^2 + v_(y)^2] = (20)^2` or `v_y = 0` Hence, it is the highest point `:. t = T/2 = (usintheta)/g` `=((20)sin60^@)/10` `=(sqrt3) s`.

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A particle is projected at an angle `60^@` with horizontal with a speed `v = 20 m//s`. Taking `g = 10m//s^2`. Find the time after which the speed of the particle remains half of its initial speed.

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