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 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. Phase 2 Rocket fuel STRATEGY Take y = 0 at the launch point and y positive upward, as in the figure. The problem consistsout of two phases. In phase 1 the rocket has a net upward acceleration of 29.4 m/s2, and we can use the kinematic equations with constant ac 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/s2, with initial velocity and position given by the results of phase 1 Apply the kinematic equations for free fall. burns Phase 1 29.4 m/s tion a to find the height Rocket crashes after falling from 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 4.00 s. (1) v = v0 + at Write the velocity and position kinematic equations. Adapt these equations to phase 1, substituting a 29.4 m/s, vo-o, and yo = o. (3) v (29.4 m/s), y-½(29,4 m/s2)r'= (14.7 m/s2)/2 (4) Substitute 1 4.00s into Equations vb = 118 m/s and y', 235 m

Explanation / Answer

a) V = at = 25.1 ( 4.05) = 101.655 m/s

h = 1/2 at^2= 205.85 m apprx

b) max height = 101.655^2/ ( 19.6) = 527. 232 m apprx

max height = 527.232 + 205.85= 733.1 m apprx

c) v = sqroot ( 2gh) = sqroot ( 2x 9.8x 733.1) = 119.869 m/s apprx

c)

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