2 AE 010. EXAMPLE 2.10 A Rocket Goes Ballistic GOAL Solve a problem involving a

Question

2 AE 010. EXAMPLE 2.10 A Rocket Goes Ballistic GOAL Solve a problem involving a powered ascent followed by free-fall motion. PROBLEM A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to Earth. (a) Find the rocket's velocity and position at the end of 4.00 s. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes on the ground. Phawe Rocket STRATEGY Take y-0 at the launch point and yu positive upward, as in the figure. The problem consists of two phases. In phase 1 the rocket has a net uward acceleration of 29.4 m/s, and we can use the kinematic equations with constant acceleration a to find the height and velocity of the rocket at the end of phase 1, when the fuel is burned up. In phase 2 the rocket is in free fall and has an acceleration of -9.80 m/s?, with initial velocity and position given by the results of phase 1. .ear-rdoaste Apply the kinematic equations for free fall. Phase 1 Two linked phases of motion for a rocket that is launched, SOLUTION (A) Phase 1: Find the rocket's velocity and position after 4.00 s. Write the velocity and (1) v=v0+ at :

Explanation / Answer

Practice it (c) velocity before crashing

v^2 = u^2 + 2gh

since it will drop from zero velocity at max height:

v^2 = 2gh = 2 x 9.8 x 1053.3 = 20644.68,

=> v = - 143.7 m/s

Exercise (Experimental rocket)

During that 132.7 meters (224– 91.3) the rocket acquires a speed of

v = (2gh) = 51 m/s

now you need the decelleration needed to slow the rocket from that speed to zero in 91.3 meters. Turning it around, it's the acceleration to reach that speed in 80m

a = ½v²/d = ½(51)²/(91.3) = 14.2 m/s²

but the rocket also has to offset the acceleration due to gravity, so a = 14.2+9.8 = 24.0 m/s²

EDIT:

Answer for "If, instead, some fuel remains, at what height should the engines be fired again to brake the rocket's fall and allow a perfectly soft landing? (Assume the same acceleration as in the initial descent.)"

At the end of 4 s its velocity is at = 29.4*4 = 117.6 m/s

The distance it has traveled up in 4 s is h = 0.5at^2 = 14.7*4^2= 235.2 m

From this point it is thrown with an initial velocity of 117.6m/s upward.

Hence it will return to this height with a downward velocity of – 117.6m/s

If at this point the engine is fired again its final velocity will be zero when it reaches the ground

Hence the rocket should be fired again when it is at a height of 235.2m

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