is stronger then the terminal velocity that is when the object is going at a small velocity. Terminal Velocity When there is equal force acting on an object when falling such as gravity and air resistance at that stage it is called constant speed or terminal velocity. When the object is dropped the force of gravity initially is 100% but as it falls the air resistance becomes stronger making the gravity weaker and at one stage there will be terminal velocity. In some cases due to the mass
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1. Alice throws the ball to the +X direction with an initial velocity 10m/s. Time elapsed during the motion is 5s‚ calculate the height that object is thrown and Vy component of the velocity after it hits the ground. 2. John kicks the ball and ball does projectile motion with an angle of 53º to horizontal. Its initial velocity is 10 m/s‚ find the maximum height it can reach‚ horizontal displacement and total time required for this motion. (sin53º=0‚ 8 and cos53º=0‚ 6) 3. The boy drops
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consideration of the forces that cause the motion. There are four activities done in this experiment. Graphical analysis of human motion‚ where displacement vs time and velocity vs. time were graphed. Graphical analysis of motion where in the 10th seconds the total displacement is 18.75m‚ average velocity is 1.88m/s and instantaneous velocity is 3.76m/s. Reaction time where one of the normal reaction time among the group is 0.16s and the reaction time while someone is distracting the member is 0.30s‚ and
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Srivastava’s Srivastava’s MOTION IN ONE DIMENSION 1. A particle is projected vertically upwards with a velocity 20 m s1 from the top of the tower of height 100 m. Determine the time it takes to reach the ground. Answer: t 2 1 6 sec. A particle is dropped from the top of the tower of height h. At the same instant another particle is projected vertically upwards from the bottom of the tower with such a velocity that it will be able to just reach the top of the tower. When and at what height from ground
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VECTOR & RELATIVE VELOCITY Set 1 1. What is the total displacement of a trip in which a person travels 10 km[N] and then 24 km[E]? 2. What is the total displacement of a trip of 50 km[W] followed by a trip of 100 km[N30°E]? 3. What is the total displacement of a trip of 100 km[N30°E] followed by a trip of 50 km[W]? What is significant about the result when compared to the answer for question 2? 4. A small boy goes to a store 2 blocks[N]‚ 3 blocks[E]‚ l block[S]‚ 5 blocks[W]
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released until it strikes the ground. 4. The time it takes for the ball to hit the ground depends on v0 ‚ g and h. 004 10.0 points The velocity of a projectile at launch has a horizontal component vh and a vertical component vv . When the projectile is at the highest point of its trajectory‚ identify the vertical and the horizontal components of its velocity and the vertical component of its acceleration. Consider air resistance to be negligible. 4. It cannot be determined. 002 10.0 points
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the displacement y of the stone? | The stone‚ starting with zero velocity at the top of the building‚ is accelerated downward by gravity. | | Reasoning The upward direction is chosen as the positive direction. The initial velocity v0 of the stone is zero‚ because the stone is dropped from rest. The acceleration due to gravity is negative‚ since it points downward in the negative direction. Solution | 2). After 3.00 s of free-fall‚ what is the velocity v of the stone? Solution 1). A football game
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at a velocity of 80 km/h when he sees an old woman crossing the road 45 m away. He immediately steps hard on the brakes to get the maximum acceleration of 7.5 m/square second. how far will he go before stopping? Will he hit the old woman? 3. the time a male bungee jumper if freely falling is 1.5 seconds (a) What is the velocity of the jumper at the end of 1.5 s? (b) how high did he fall? 4. A juggler tosses three balls alternately vertically upward. each ball has an initial velocity of 5
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30’+60’ = 90’=5400s=1.5h 40km+40km=80km vavg= 80000m/5400s=13.168724279835390946502057613169 m/s = 48 km/h 2. The coordinate of an object is given as a function of time by x = 7t – 3t2‚ where x is in meters and t is in seconds. Its average velocity over the interval from t = 0 to t = 2 s is: A) 5 m/s B) –5 m/s C) 11 m/s D) –11 m/s E) –14.5 m/s vt= 7-6t = 7-12=-5 m/s 3. The coordinate of a particle in meters is given by x(t) = 16t – 3.0t3‚ where the time t is in seconds. The
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1 hour IB DIPLOMA PROGRAMME PROGRAMME DU DIPLÔME DU BI PROGRAMA DEL DIPLOMA DEL BI N07/4/PHYSI/HPM/ENG/TZ0/XX+ 88076501 Physics higher level PaPer 1 Thursday 8 November 2007 (afternoon) INSTRUCTIONS TO CANDIDATES • Do not open this examination paper until instructed to do so. • Answer all the questions. • For each question‚ choose the answer you consider to be the best and indicate your choice on the answer sheet provided. 8807-6501 21 pages © IBO 2007 –2– 1. Which
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