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With a vertical velocity of 2.9m per second and a horizontal velocity of 8.3m per second what is the angle of takeoff?
Wiki User
∙ 16y ago
About 19.24 degrees, assuming no change of velocity in either direction. Draw two vectors. A horizontal vector of length 8.3. and a vertical vector of length 2.9 and connect the "tail" of the second to the "head" or "arrow" of the first. You should have a horizontal line segment of 8.3 and a vertical line segment of 2.9 rising from the right end of the horizontal line segment. You'll have two sides of a right triangle, with the right angle on your right. Connect the ends of the two segments with another segment, and that's the hypotenuse of your right triangle. You're interested in the angle on the left - the takeoff angle. You know the length of the side opposite it and the length of the side adjacent to it. The tangent function is opposite over adjacent. tan = opp / adj You need the angle whose tangent is found by dividing the length of the opposite side by the length of the adjacent side. When we see "the angle whose tangent is" we use arctangent. We'll use T as the takeoff angle. arctan T = 2.9 / 8.3 = .349 arctan of .349 = 19.24o
Wiki User
∙ 16y ago
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When naming a pair of coordinates which?
First: horizontal: Abscissa Second: vertical: ordinate.
If a ball rolls off the edge of a table two meters above the floor and with an initial velocity of 20 meters per second what is the ball's acceleration and velocity just before it hits the ground?
The horizontal velocity has no bearing on the time it takes for the ball to fall to the floor and, ignoring the effects of air resistance, will not change throughout the ball's fall, so you know Vx. The vertical velocity right before impact is easily calculated using the standard formula: d - d0 = V0t + [1/2]at2. For this problem, let's assume the floor represents zero height, so the initial height, d0, is 2. Further, substitute -g for a and assume an initial vertical velocity of zero, which changes our equation to 0 - 2 = 0t - [1/2]gt2. Now, solve for t. That gives you the time it takes for the ball to hit the floor. If you divide the distance traveled by that time, you know the average vertical velocity of the ball. Double that, and you have the final vertical velocity! (Do you know why?) Now do the vector addition of the vertical velocity and the horizontal velocity. Remember, the vertical velocity is negative!
The vertical axis in a coordinate?
Horizontal and vertical axes are alphabetical, (x,y), abscissa and ordinate, first and second, and so forth. The only place I see it not alphabetical is independent and dependent.
If raindrops fall vertically at a speed of 3 meters per second and you are running at 4 meters per second how fast do they hit your face?
3m/s vertical velocity and 4m/s horizontal velocity using vector addition you can add both vectors with Pythagoras a^2 + b^2 = c^2 3^2 + 4^2 = c^2 9 + 16 = c^2 c = 5 m/s you can use trigonometry to find the angle of the velocity vector also
Is y equals 2x-2 and y equals 2 parallel?
No. The slope of the first is 2 - ie a change in the horizontal direction results in double the change in the vertical direction. The second line is horizontal (slope = 0).
A banana is thrown straight out so it has both horizontal and vertical velocity After 1 second, what is its vertical velocity?
9.8
How does the unbalanced force of gravity affect the horizontal and vertical velocity of an object in projectile?
In the absence of air resistance, the force of gravity has no effect on the horizontal component of a projectile's velocity, and causes the vertical component of its velocity to increase by 9.8 meters (32.2 feet) per second downward for every second of its flight.
How does the unbalanced force of gravity affect the horizontal and vertical velocities of an object in projectile?
In the absence of air resistance, the force of gravity has no effect on the horizontal component of a projectile's velocity, and causes the vertical component of its velocity to increase by 9.8 meters (32.2 feet) per second downward for every second of its flight.
What is the magnitude of the velocity of a vertical projectile at its maximum height is equal to?
The horizontal component of a projectile's velocity doesn't change, until the projectile hits somethingor falls to the ground.The vertical component of a projectile's velocity becomes [9.8 meters per second downward] greatereach second. At the maximum height of its trajectory, the projectile's velocity is zero. That's the pointwhere the velocity transitions from upward to downward.
When a stone is thrown upward at an angle what happens to the vertical component of its velocity as it rises and as it falls?
The vertical component of its velocity increases at the rate of 9.8 meters (32.2 feet) per second downward every second. Without involving numbers, simply the vertical component will first be upward at what ever velocity it is when split from the horizontal velocity, then (after reaching the peak of its height at which velocity is zero) it will be a downward vector that, yes, will increase with acceleration due to gravity (which is where the 9.8 meters per second squared came from)
A ball of mass m rolls off a cliff of height h with a horizontal velocity v it reaches the ground with vertical speed of?
If one assumes air resistance to be negligible, then: final velocity = sqrt( g * 2 * h ) where g is 9.8 metres per second per second. The quantities v and m do not matter, because gravitational acceleration does not depend on mass (all objects fall at the same rate) and because the horizontal velocity is independent of the vertical velocity.
A projectile is fired horizontally from a gun that is 71.0 m above flat ground emerging from the gun with a speed of 250 ms What is the magnitude of the vertical component of its velocity as it st?
If it's fired horizontally, then its initial vertical velocity is zero. After that, the vertical velocityincreases by 9.8 meters per second every second, directed downward, and the projectile hitsthe ground after roughly 3.8 seconds.Exactly the same vertical motion as if it were dropped from the gun muzzle, with no horizontal velocity.
Can stress be resolved into horizontal and vertical components?
Force can be resolved into horizontal and vertical components using vector analysis. However stress cannot be resolved into horizontal and vertical components using vector analysis since it is not a vector but a tensor of second order.
A ball is thrown horizontally from a cliff at a speed of 8 meters per second. What is its speed one second later?
After a second, the ball will still have a horizontal velocity of 8 meters per second. It will also have a vertical velocity of 9.8 meters per second (Earth's acceleration is about 9.8 meters per square second). The combined speed (using the Law of Pythagoras) is about 12.65 meters per second.
When naming a pair of coordinates which?
First: horizontal: Abscissa Second: vertical: ordinate.
If a ball rolls off the edge of a table two meters above the floor and with an initial velocity of 20 meters per second what is the ball's acceleration and velocity just before it hits the ground?
The horizontal velocity has no bearing on the time it takes for the ball to fall to the floor and, ignoring the effects of air resistance, will not change throughout the ball's fall, so you know Vx. The vertical velocity right before impact is easily calculated using the standard formula: d - d0 = V0t + [1/2]at2. For this problem, let's assume the floor represents zero height, so the initial height, d0, is 2. Further, substitute -g for a and assume an initial vertical velocity of zero, which changes our equation to 0 - 2 = 0t - [1/2]gt2. Now, solve for t. That gives you the time it takes for the ball to hit the floor. If you divide the distance traveled by that time, you know the average vertical velocity of the ball. Double that, and you have the final vertical velocity! (Do you know why?) Now do the vector addition of the vertical velocity and the horizontal velocity. Remember, the vertical velocity is negative!
How do you draw a horizontal?
It's parabolic. More specifically, it's half a parabola, with a vertex at the point of origin. You see, horizontal and vertical velocity are independent of each other. This means that the projectile will more horizontally at a constant velocity (ignoring wind resistance) equal to its initial velocity. At the same time, the projectile is accelerated downward at a rate of 9.8 m/s2 (approximately). Since it undergoes constant acceleration, it increases in velocity every second exponentially, giving it a parabolic curve. As an interesting side note, since horizontal and vertical velocity are always independent of each other, any velocity (really, any vector) can be represented as its horizontal and vertical components, allowing one to add up the individual components and to find a resultant vector quantity. Here's an example drawing of the curve described: http://www.staff.amu.edu.pl/~romangoc/graphics/M2/1-projectile-motion/M2-3.gif Keep in mind that if the point of origin is at a greater vertical level than the ground, the projectile will fall all the way to the ground, past the horizontal axis.
