A classical mechanics problem by Manas Wadhwa

For bullet to hit the plane it must travel the same distance in the horizontal direction as the plane.

Let $t$ be the time after the bullet hits the plane.

As the velocity of plane is in horizontal direction therefore the distance covered is $200t$ .

The distance travelled by the bullet in horizontal direction when shot with an angle $\alpha$ can be given by $400(\cos{\alpha})t$ .

Equating both we get.

$400(\cos{\alpha})t=200t\\\cos{\alpha}=\dfrac{1}{2} \\ \alpha=60^{o}$ .

SO GIVEN AB=1500m Let BC=x1 AC=x2 Vb=400m/s v=s/t t=x1/400 Va=200m/s v=s/t t=x2/200 t/t=x1/2x2 1*2x2=x1 2x2=x1 equation 1 pythagoras theorem (1500)^2+(x^2)^2=(x1)^2 put x1=2x2 x1=500 √5 x2=1000√5 now sin theta =AC/BC theta =30 degree alpha = 90-theta = 60degrees