Question

A plane is flying horizontally at a height 1.5 km from the ground with a velocity 200m/s when it is just above a anti airgun at that moment a bullet is thrown from it with velocity400m/s.find the angle of projection of bullet to crush the plane?

Answer 1

`60°`

Explanation:

`"Name"`
`v_0 = " initial velocity of the bullet " = 400 m/s`
`v_{0x} = " horizontal initial velocity of the bullet"`
`v_{0y} = " vertical initial velocity of the bullet"`
`t = " time in seconds between firing the bullet and impact"`
`x = " horizontal distance in meter between place of firing and"`
`"place of impact"`

`"Then we have"`
`x = 200 * t`
`x = v_{0x} * t`
`=> v_{0x} = 200`

`v_{0x}^2 + v_{0y}^2 = 400^2 = 160000`
`=> v_{0y}^2 = 160000 - 200^2 = 160000 - 40000 = 120000`
`=> v_{0y} = 200 sqrt(3)`

`tan(theta) = v_{0y} / v_{0x} = sqrt(3)`
`=> theta = 60°`

`"So we do not need the given that the plane is flying at an"`
`"altitude of 1.5 km."`