It actually can be seen - velocity vector is completely horizontal.Ģ) in yellow scenario, the angle is smaller than the angle in the first (red) scenario. This means that the horizontal component is equal to actual velocity vector. In this case, this assumption (identical magnitude of velocity vector) is correct and is the one that Sal makes, too).ġ) in blue scenario, the angle is zero hence, cosine=1. So from our derived equation (horizontal component = cosine * velocity vector) we get that the higher the value of cosine, the higher the value of horizontal component ( important note: this works provided that velocity vector has the same magnitude. If we work with angles which are less than 90 degrees, then we can infer from unit circle that the smaller the angle, the higher the value of its cosine. Now we get back to our observations about the magnitudes of the angles. Horizontal component = cosine * velocity vector After manipulating it, we get something that explains everything! We do this by using cosine function: cosine = horizontal component / velocity vector. If above described makes sense, now we turn to finding velocity component. 66 11.6 (0.66 ) 7.7 m m yi s s m m ss m x s y v t a t t t t s x v t x s m building vx vyi v 50° y x vyi = 13.8 ms a = g = -10 2 m s vx = 11.6 ms x = 24 m t = ? y = ? vyi = 13.8 ms a = g = -10 2 m s vx = 11.6 ms y = 0 t = ? x = ? vyi = 13.8 ms a = g = -10 2 m s vx = 11.6 ms y = 6.All thanks to the angle and trigonometry magic.Īfter looking at the angle between actual velocity vector and the horizontal component of this velocity vector, we can state that:ġ) in the second (blue) scenario this angle is zero Ģ) in the third (yellow) scenario this angle is smaller than in the first scenario. The balloon can be launched from less than 24 m away from the building at the same speed and angle and still hit exactly the same height you calculated in part a. If the balloon misses or shoots over the building, how far will the balloon land from its launch location? 2 2 2 21 1 2 2 0 (13.8 ) ( 10 ) (5 ) 13.8 2.8 (11.6 )2.8 32 m m yi s s m m ss m x s y v t a t t t t t s x v t s m c. A water balloon is launched at a building 24 m away with an initial velocity of 18 m/s at an angle of 50˚ above the horizontal. The bottom of the ramp is 0.90 m above the floor. Now the ramp is tilted downwards and the sphere leaves the ramp at 1.5 m/s as shown below. The end of the ramp is 1.20 m above the floor. A metal sphere is launched with an initial velocity of 1.5 m/s as it leaves the ramp. Download Particle Motion in Two Dimensions Model Worksheet 4: Projectile Motion Problems and more Physics Exercises in PDF only on Docsity!1 U6 2D Motion - ws 4 v3.1 Name Date Pd Particle Motion in Two Dimensions Model Worksheet 4: Projectile Motion Problems 1.
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