Example 2.10 A Rocket Goes Ballistic Goal Solve a problem involving a powered as

Question

Example 2.10 A Rocket Goes Ballistic Goal Solve a problem involving a powered ascent followed by free fall motion Maximum height ymax 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 5.16 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 5.16 s (b) Find the maximum height the rocket reaches Phase 2 a =-9.80 m/s2 (c) Find the velocity the instant before the rocket crashes on the ground Strategy Take y = 0 at the launch point and y positive upward, as in Figure 2.21. The problem consists of two phases. In phase 1, the rocket has a net upward 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. Apply the kinematic equations for free fall Rocket fuel burns out Phase 1 41 Rocket crashes after falling from max Launch Figure 2.21 Two linked phases of motion for a rocket that is launched, uses up its fuel, and crashes Solution (a) Phase 1: Find the rocket's velocity and position after 5.16 s Write the velocity and position kinematic equation:s

Explanation / Answer

distance travelled before the engine starts is 265-84.4 = 180.6 m

speed of the rocket before the rocket starts is Vo = sqrt(2*g*h) = sqrt(2*9.81*180.6) = 59.5 m/s

final speed is V = 0

distance travelled before reaching the ground from the point where the engine is started is S = 84.4 m

accelaration is a

then using

V^2 - Vo^2 = 2*a*S

0^2 - 59.5^2 = 2*a*84.4

accelaration is a = -20.97 m/s^2

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distance travelled by rocket fall before starting to slow down is S = 180.6 m

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