Question

Please do step by step clear and show eqations used. Thank you

A rocket designed to place satellites into orbit is carried to a cruising altitude. When the plane is flying in a straight line and at a constant speed of 850 km/h, the rocket is dropped. After the drop the plane continues to fly at the same pseed and in a straight line. The rocket falls for a bried time, and then its propulsion starts. Once the propulsion starts the combined effects of gravity and thrust give the rocket a non-constant acceleration of (29.43 + 0.44t) m/s^2 directed at an angle of 30.0 degrees above the horizontal. For safety the rocket should be at least 1.00km in front of the plane when it climbs through the plane's altitude. Ignore drag.

a) If the rocket falls for 5.15 s before engaging propulsion, how much total time will pass before the rocket passes back through the cruising altitude of the plane?
b) will the rocket safely pass 1.oo km in front of the plane for this free-fall time of 5.15s? What will be the horizontal distance between the rocket and the plane when they are at the same altitude ?

Answer 1

Let as assign time = 0 at the moment when propulsion of the rocket starts. Hence the time when it was released from the plane becomes time = -5.15 s.

When the rocket got separated from the plane there was no change in the velocity of the plane. This implies the velocity of the rocket just after the separation must also be equal to the velocity of the plane which is 850 km/hr = 236.11 m/s in horizontal direction. This is from the law of conservation of linear momentum.

The rocket then freely falls under gravity for 5.15 seconds. During this fall the plane and the rocket will always be in the same vertical line as there horizontal velocities are same. The vertical distance covered by rocket during this time can be calculated as:

= 0.5 x 9.8 x (5.15)²

= 130 m

and velocity of the rocket in downward direction:

u_ry = at

= -9.8 x 5.15

u_ry = -50.47 m/s (Taking upward as +ve y direction)

When the rocket has fallen for this 130 m on 5.15 s time, now the propulsion in it starts. The rocket now accelerates with acceleration a = (29.43 + 0.44t) m/s^2 at an angle 30^o with the horizontal. Resolving ths acceleration for horizontal and vertical direction.

a_x = (29.43 + 0.44t)cos30^o

a_x = (25.48 + 0.28t) m/s²

a_y = (29.43 + 0.44t)sin30^o

a_y = (14.71 + 0.22t) m/s²

Let us assume the rocket reaches the altitude of the plane at time = t seconds. The verical distance travelled by the rocket during this time will be equal to 130 m.

For vertical direction:

The vertical distance travelled by the rocket will be given by

  ANS (a)

For Horizontal direction:

Distance travelled in horizontal direction during the the time 8.56 seconds when the rocket reaches the level of the plane:

= 292.91 + 933.5 + 2021.1 m

= 3247.51 m

During this 8.56 seconds the plane will also move in the horizontal direction by

8.56 x 236.11 = 2021.1m

Hence the rocket will be ahead of the plane by a distance of

3247.51 - 2021.1 = 1226.41m = 1.22641 km

The rocket will be 1.22641 km ahead of the plane which is more than the minimum safe distance (1km) required.