Particle projected from tower fo heigh 10m as shown in figure. Find the time (in sec) after which particle will hit the ground.

Answer

2

Explanation

`u_(y)=v sin 45^(@)=5sqrt(2)xx(1)/sqrt(2)=5ms^(-1)`
Using, `y=u_(y)t+(1)/(2)a_(y)t^(2)`
`-10=5t+(1)/(2)(-10)t^(2)`
`t^(2)-t=2=0`
`(r-2)(t+1)=0`
`therefore` t=2 (rejecting -ve time)