A particle `A` is projected from the ground with an initial velocity of `10m//s` at an angle of `60^(@)` with horizontal. From what height should an another particle `B` be projected horizontally with velocity `5m//s` so that both the particles collide in ground at point `C` if both are projected simultaneously `g=10 m//s^(2)`.

Options

A. 10 m
B. 15 m
C. 20 m
D. 30 m

Answer

B

Explanation

Both the particles will meet at C, if their time of flight is the same. The time of flight of A is
`T=(2usintheta)/g=(2xx10 xx sin60^@)/10 =sqrt3s`
For vertical downward motion of particle
`h=1/2 gT^2 =1/2 xx10 xx(sqrt3)^2=15m`