Question

Question #1e5b9

Answer 1

Relative velocity of projectile w.r.t plane=Actual velocity of projectile-velocity of plane
So,Actual velocity of projectile= `(300+500*(5/18))`m/sec or, 438.89 m/sec
Let's,consider that the projectile will touch the ground after time t,
So with in this time,along X axis if it travels a distance of s from its point of projection,we can say
`s=(438.89)*t`....1

And along Y axis,
Vertical height of plane above the ground = 800 = `(1/2)g(t^2)`...2

Eliminating t from both the equations we get,
800 = `(1/2)10(s/438.89)^2`
Or, s= 5551.56 meters
So at that time along x axis its velocity will be 438.89 m/sec
Along Y axis we can find it using equation `v^2 = u^2 +2as`(all symbols are bearing their conventional meaning)
So,v= 126.5 m/sec
Hence net velocity with which it will hit the ground will be
`√((Vx)^2 + ((Vy)^2))`
Or, 456.75 m/sec