A particle is projected from a tower of height 25 m with velocity `20sqrt(2)m//s` at `45^@`. Find the time when particle strikes with ground. The horizontal distance from the foot of tower where it strikes. Also find the velocity at the time of collision.

Answer

A, B, C, D

Explanation

`u_x = u_y = 20 m//s, a_y = -10 m//s^2`
`s_y = u_yt + 1/2 a_yt^2`
`rArr -25 = 20t - 1/2 xx 10 xx t^2`
Solving this equation, we get the positive value of ,
`t= 5 sec`
Now apply, `v=u + at and s_x = u_xt` .