# Question 1e5b9

Jan 13, 2018

Relative velocity of projectile w.r.t plane=Actual velocity of projectile-velocity of plane
So,Actual velocity of projectile= $\left(300 + 500 \cdot \left(\frac{5}{18}\right)\right)$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
$s = \left(438.89\right) \cdot t$....1

And along Y axis,
Vertical height of plane above the ground = 800 = $\left(\frac{1}{2}\right) g \left({t}^{2}\right)$...2

Eliminating t from both the equations we get,
800 = $\left(\frac{1}{2}\right) 10 {\left(\frac{s}{438.89}\right)}^{2}$
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 ${v}^{2} = {u}^{2} + 2 a s$(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