Question

An aircraft of mass M =1.20×104 kg is practicing a bombing run on a tunnel on Earth that slants forward and down into the ground at an angle of θ =55.0° from the horizontal as shown in the figure. The aircraft is constrained to fly at a constant height of h above the ground at a constant speed of v0 = 750 km/h. Upon release, the bomb (mass m =150kg) executes free fall with constant downward acceleration of g=9.8m/s2. Air resistance is negligible. The bomb must enter with its velocity parallel to the tunnel. (a) Find the height h in km at which the aircraft must fly so that the bomb enters the tunnel at the correct slope. (b) Find the distance a (in km) from the tunnel at which the bomb must be released in order to hit the tunnel. (c) Find the speed of the bomb entering the tunnel in km/h

Answer 1

Here,

theta = 55 degree

mass , M = 1.20 *10^4 kg

constant speed , v0 = 750 km/hr

v0 = 208.33 m/s

let the vertical velocity is vy

tan(theta) = vy/v0

tan(55) = vy/208.33

vy = 297.53 m/s

let the height is h

h = vy^2/(2 * g)

h = 297.53^2/(2 * 9.8)

h = 4516 m

the height h of the aircraft is 4516 m

b)

time of flight , t = vy/g

t = 297.53/9.8 s

t = 30.4 s

horizontal distance , d = 208.33 * 30.4 m

horizontal distance , d = 6325 m

the distance from the tunnel is 6.32 km

c)

speed of bomb = sqrt(vo^2 + vy^2)

speed of bomb = sqrt(297.53^2 + 208.33^2 )

speed of bomb =363.2 m/s = 1307 km/hr

the speed of bomb is 1307 km/hr