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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?
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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
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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.
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?
The vertical component of velocity for the projectile when it is fired horizontally is zero. This is because the initial velocity is entirely in the horizontal direction, and there is no initial velocity in the vertical direction. Gravity will act on the projectile, causing its vertical velocity to increase as it travels.
A ball is thrown horizontally from a window on the second floor of a building What is the vertical component of its initial velocity?
The vertical component of the initial velocity of the ball thrown horizontally from a window is zero. The ball's initial velocity in the vertical direction is influenced only by the force of gravity, not the horizontal throw.
How does the unbalanced force of gravity affect the horizontal and vertical velocities of an object in projectile?
The unbalanced force of gravity affects only the vertical velocity of an object in projectile motion. Gravity acts downward to accelerate the object vertically, while the horizontal velocity remains constant in the absence of other forces. This results in a curved path known as a projectile trajectory.
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?
As the stone rises, the vertical component of its velocity decreases due to the acceleration of gravity acting in the opposite direction of the initial velocity. When the stone falls, the vertical component of its velocity increases in the negative direction as gravity accelerates it downwards.
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.
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.
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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.
When naming a pair of coordinates which?
First: horizontal: Abscissa Second: vertical: ordinate.
What are the two types of motion a projectile has?
If you throw ball at an angle above horizontal, you will see the path of the ball looks like an inverted parabola. This is result of the fact that the ball's initial velocity has a horizontal and vertical component. If we neglect the effect of air resistance, the horizontal component is constant. But the vertical component is always decreasing at the rate of 9.8 m/s each second. To illustrate this, let the initial velocity be 49 m/s and the initial angle be 30˚. Horizontal component = 49 * cos 30, Vertical = 49 * sin 30 = 24.5 m/s As the ball rises from the ground to its maximum height, its vertical velocity decreases from 24.5 m/s to 0 m/s. As the ball falls from its maximum height to the ground, its vertical velocity decreases from 0 m/s to -24.5 m/s. Since the distance it rises is equal to the distance it falls, the time that it is rising is equal to the time it is falling. This means the total time is equal to twice the time it is falling. This is the reason that the shape of the ball's path is an inverted parabola. At the maximum height, the ball is moving horizontally. If you do a web search for projectile motion, you will see graphs illustrating this.
