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During Oblique Impact does the vertical component of velocity change after impact?
Wiki User
∙ 13y ago
When a ball collides with a wall, at an angle θ say, the impulse exerted on the ball is perpendicular to the wall and causes a change in the momentum of the ball in that direction; it does not however affect the momentum parallel to the wall.
Therefore, if the approach velocity of the sphere is resolved into components parallel and perpendicular to the wall, one of these components is changed by the impact and the other remains unchanged.
Wiki User
∙ 13y ago
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A projectile fired from the ground follows a parabolic path The speed of the projectile is minimum at the top of its path?
Yep that is correct. To understand this it may help you to draw a parabola. If you draw a line from the top of the parabola back to the ground you'd notice either side of this line is symmetrical. This isn't quite what happens to a projectile (following a parabolic path), but because of the nature of the question, effects such as air resistance can be ignored. As the projectile approaches the top of its path, the vertical component of its velocity approaches zero. As the projectile begins to fall the magnitude of the vertical component of the projectile begins to increase. The only force that acts on the projectile during flight is gravity which pulls it towards the earth. Since this force and the horizontal component of the projectiles velocity are at right angles to each other, the horizontal component of the velocity is unaffected during flight . This explains the symmetry of the parabola and also means the time to reach the top of path equals the time from the top of path back to the ground. The projectile will hit the ground with the same speed as it left the ground. If you draw a horizontal line through the parabola, at the two points where the line and the parabola cross, the speed of the projectile will be the same. The only change to the balls speed during the flight comes as the vertical component of its velocity tends to zero as it reaches the top of the curve and then falling back down due to gravity. I'm unsure of your physics knowledge but hopefully this doesn't confuse you. If you have learned about vectors, then this can be simply understood/explained.
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The horizontal and vertical components of a projectile's velocity are independent of each other what does this statement mean?
That means the if you change one you do not necessarily change the other. In the case of the projectile the vertical component is dependent on time (if it is a projectile near a large mass like the earth) gravity acts on it accelerating the projectile in a downward direction. The horizontal component remains the same during the entire flight (if we disregard air resistance and such things).
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A projectile fired from the ground follows a parabolic path The speed of the projectile is minimum at the top of its path?
Yep that is correct. To understand this it may help you to draw a parabola. If you draw a line from the top of the parabola back to the ground you'd notice either side of this line is symmetrical. This isn't quite what happens to a projectile (following a parabolic path), but because of the nature of the question, effects such as air resistance can be ignored. As the projectile approaches the top of its path, the vertical component of its velocity approaches zero. As the projectile begins to fall the magnitude of the vertical component of the projectile begins to increase. The only force that acts on the projectile during flight is gravity which pulls it towards the earth. Since this force and the horizontal component of the projectiles velocity are at right angles to each other, the horizontal component of the velocity is unaffected during flight . This explains the symmetry of the parabola and also means the time to reach the top of path equals the time from the top of path back to the ground. The projectile will hit the ground with the same speed as it left the ground. If you draw a horizontal line through the parabola, at the two points where the line and the parabola cross, the speed of the projectile will be the same. The only change to the balls speed during the flight comes as the vertical component of its velocity tends to zero as it reaches the top of the curve and then falling back down due to gravity. I'm unsure of your physics knowledge but hopefully this doesn't confuse you. If you have learned about vectors, then this can be simply understood/explained.
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If "range" means that the shooter and the target are on the same level: quadrupled (if airesistance can be neglected). It takes twice the time until gravity "eats up" vertical velocity and during that time the projectile moves with double horisontal velocity. But if you shoot horisontally from a cliff at double velocity the flighttime will be the same and the range only doubled.
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What is the difference between group velocity and phase velocity in a stationary wave on a string?
The velocity of a wave which maintains consatnt phase at all successive positions during propogation is known as wave velocity or phase velocity. The velocity of a group of waves which maintains constant poditions during the propogation is known as group velocity.
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Force = mass * acceleration Mass is only involved during the acceleration in the gun barrel , and is involved (with the explosive force) in translating to muzzle velocity. The horizontal distance travelled depends on the muzzle velocity and the incline of the barrel to horizontal. The curve will be parabolic even when the launch angle is 0 in which case the path will be negative (essentially going underground) 1. split launch velocity into horizontal and vertical vectors 2. using vertical velocity vector (initial velocity u), calculate (total) time to rise and fall back to ground using newtons equations. 3. multiply time by horizontal velocity vector to calculate horizontal distance travelled to landing site.
