The horizontal and vertical distances covered by a projectile at tiem t are given by x=at and y=`bt^(2)+ct`, wehre a,b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?

Answer

`sqrt(a^(2)+(2b+c)^(2))`

Explanation

`v_(x)=(dx)/(dt)=(d)/(dt)(at)propa`
`v_(y)=(dy)/(dt)=(d)/(dt)(bt^(2)+ct)=2bt+c`
At t=1s, `u_(y)=2b+c`
The magnitude of velocity at t=1s is given by
`v=sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(a^(2)+(2b+c)^(2))`