A catapult launches a test rocket vertically upward from a well, giving the rock

Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level. The engines then fire, and the rocket accelerates upward at 4.20 m/s2 until it reaches an altitude of 960 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of ?9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)

(a) For what time interval is the rocket in motion above the ground?




(b) What is its maximum altitude?





(c) What is its velocity just before it hits the ground?

Explanation / Answer

rocket has an initial speed (at ground level) = 80 m/s rocket has an upward acceleration at ground level = 4 m/s² until 1000 m. d = Vo(t) + 1/2at² {where d = 1000, Vo = 80, a = 4, find t} 0 = 2t² + 80t - 1000 solve for pos root of t: t = 10 s {time for rocket to reach 1000 m after getting to ground level} rocket height = h above 1000m: h = Vi(t) - 1/2gt² Vi = v(Vo)²+2ad = v(80)²+2(4)(1000) = v6400+8000 = v14,400 = 120 m/s time to reach max height after reaching 1000 m = 120/9.80 = 12.24s h = 120(12.24) - (0.5)(9.80)(12.24)² = 1468.8 - 734.1 = 734.7 ˜ 735 m rocket's max height above ground = 1000 + 735 = 1735 m ANS b) time to free-fall 1735 m = v2(1735)/9.80 = v354.0816 = 18.8 s TOTAL TIME interval above ground = 10.0 + 12.2 + 18.8 = 41.0 s ANS a) VELOCITY just before rocket hits ground = v2gh = v(2)(9.80)(1735) = 184 m/s ANS c)

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