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?
s

(b) What is its maximum altitude?
km

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

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)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.