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Unformatted text preview: homework 03 RAHMAN, TARIQUE Due: Feb 4 2008, 3:00 am 1 Question 1, chap 2, sect 6. part 1 of 1 10 points A toy rocket, launched from the ground, rises vertically with an acceleration of 20 m / s 2 for 9 . 3 s until its motor stops. The acceleration of gravity is 9 . 8 m / s 2 . Disregarding any air resistance, what max imum height above the ground will the rocket achieve? Correct answer: 2 . 63 km (tolerance 1 %). Explanation: Basic Concepts: v = v + at y = v t + 1 2 at 2 since the initial position is zero. Solution: While the motor is running, the rocket rises to a height of h 1 = 1 2 a t 2 = 1 2 ( 20 m / s 2 ) ( t ) 2 1 km 1000 m = 0 . 8649 km When the motor stops running, the speed of the rocket is v 1 = at At this point in time the rocket continues to rise, but now its acceleration (deceleration) is due to gravity alone, and its initial velocity is v 1 = at . At its maximum height its velocity is zero, so v 2 = 0 = v 2 1 2 g h 2 and 2 gh 2 = v 2 1 h 2 = v 2 1 2 g = ( at ) 2 2 g = ( 20 m / s 2 ) 2 (9 . 3 s) 2 2 (9 . 8 m / s 2 ) = 1 . 7651 km Thus the total height obtained by the rocket is h = h 1 + h 2 = 0 . 8649 km + 1 . 7651 km = 2 . 63 km Question 2, chap 2, sect 7. part 1 of 1 10 points An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the acceleration a of the object is given by a = g b v , where v is the objects speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time? 1. v = ( g + a ) t b 2. v = ( g e bt ) b 3. v = v + g t, v negationslash = 0 4. v = g t b t 2 5. v = g ( 1 e bt ) b correct Explanation: At time t = 0, the speed of the object is zero, and at time t = , the acceleration is zero, corresponding to a speed v = g b . Check the five choices, and it shows that the only possible answer is v = g ( 1 e bt ) b . Note: The answer can be directly obtained by integration: a = g b v d v dt = g parenleftbigg b v g 1 parenrightbigg homework 03 RAHMAN, TARIQUE Due: Feb 4 2008, 3:00 am 2 integraldisplay v dv b g v 1 = g integraldisplay t dt g b ln parenleftbigg b g v 1 parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle v = g t vextendsingle vextendsingle vextendsingle vextendsingle t g b bracketleftBig ln parenleftbigg b g v 1 parenrightbigg ln( 1) bracketrightBig = g t ln parenleftbigg 1 b g v parenrightbigg = b t 1 b g v = e b t b g v = 1 e b t v = g b parenleftBig 1 e b t parenrightBig . Question 3, chap 2, sect 7....
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 Fall '08
 Turner
 Physics, Acceleration, Gravity, Work

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