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Activity in Kinematics

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  Activity in Kinematics Part I. Distance (in) Time (s) Velocity (in/s) 80 3.29 24.31610942 95.5 4.38 21.88449848 A toy car slows down from 24.32 in/s to 21.88 in/s. What is the acceleration of the toy car? in m/s? Given:   = 24.31610942 in/s    = 21.88449848 in/s    = 3.29 s    = 4.38 s Required: a.)   a Solutions: a.) = ∆∆  =        −    = .444 in  − 4.364 in  4.3 s − 3. s  = −.4364  .      = -2.230835725  /    ~ -2.23        b.) Acceleration in m/s Given that there are 0.0254 m in 1 inch, convert -2.230835725  /    to m/s -2.230835725     ˟ .54    = -0.05666322741    ~ -0.0567    Part II. A girl drops a ball to the ground as she travelled 39 inches at a time 0.41 seconds. The height of the ball prior to the ground is 40 inches. How long is the ball in the air? How far does the ball travel horizontally? Given: 1.) Description Distance (in) Time (s) Velocity (in/s) Velocity before it was dropped 39 0.41 95.12195122   Description Distance (in)   Time (s)   Velocity (  Initial Velocity (Horizontal Component)   39   0.41   95.12195 Vertical Components   40   0.41   2.) h = 40 in  We measured the height of the ball and the runner makes sure that the height of the ball from the ground will be constant as she run at a constant velocity of 59.58 in/s straight. Anyway, the average constant velocity of the runner does not affect the mechanics of the ball; only the instantaneous velocity which is the initial velocity of the ball in horizontal dimension or the velocity before she drops the ball acted upon  by the acceleration due to gravity. Keeping the height of the ball constant, makes sure that the ball will just be projected horizontally  before it will hit the ground. Solutions: a.)      =    +      =        Note:       has a zero value since it is under the vertical component of our horizontally projected ball and the value is independent from the mechanics of the horizontal components.    Convert first the 0 in/s to m/s and the result is obviously 0 as well in m/s. Also, convert 40 in to m. 40     ˟ .54    = 1.016 m       has a negative value since our ball is moving downward.    Find first the    so we can solve for the time. (  )   - (  )   = 2gh (  )   = (  )   + 2gh (−  )   −   = (  )   + gh−    (  )   = (  )   + gh−  = ( /)   + (−.665  )(.6 m)−  = (   /  ) + (−.   /  )−  = −.   /  −  = 19.9271128     /    (  )   = 19.9271128     /       = √19.9271128   /       = 4.46397948 m/s  Now, solve for the time the ball was in the air.   = −      = −4.4634   −   −.665   = −4.4634  −.665   = 0.455199225 s ~ 0.46 s  b.)   How far does the ball travel horizontally?  Note:    This is from the point where the lady dropped the ball to the  point where the ball hit the ground.    In this case, we will just be using the horizontal components only.    Convert first the given initial velocity of the ball = 95.12195122 in/s to m/s.
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