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What is total time hang time of a projectile?
The total hang time of a projectile is the time it takes for the projectile to go up and come back down to its original height. It can be calculated using the formula: Total hang time = 2 * time to reach maximum height.
What is the initial velocity of a bullet fired verticaly upward from a gun and reaches a height of 7000 ft?
The initial velocity of the bullet can be obtained by using the kinematic equation for projectile motion. Assuming we neglect air resistance, the initial velocity of the bullet fired vertically upward from a gun can be calculated by setting the final velocity as 0 when it reaches the maximum height of 7000 ft. Using the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity, a is the acceleration due to gravity, and s is the total displacement. Solve for u to find the initial velocity of the bullet.
What happens to the total energy of the pendulums as it swings What determines the maximum total energy of the pendulum?
As the pendulum swings, the total energy (kinetic + potential) remains constant if we ignore friction. The maximum total energy of the pendulum is determined by the initial conditions such as the height from which it is released and the velocity. The higher the release point and the greater the initial velocity, the higher the maximum total energy of the pendulum.
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.
What quantities remain constant in projectile motion?
All that I can think of are: 1.) Gravity 2.) Wind 2.A) wind speed 2.B) direction of wind 3.) Angle of trajectory 4.) Initial speed of projectile 5.) Material through which projectile travels (as in density) 6.) Mass of projectile 7.) Spin 7.A) speed of spin 7.B) axis/axes spining occurs on 8.) Shape of projectile 9.) Temperature of medium projectile is in 10.) Size of projectile (as in height, width, and depth) 11.) Weighting of projectile 12.) Obsturctions to projectile's path In a vaccuum, though, these are the variables: 1.) Speed of object 2.) Obstructions in path 3.) Gravity
What is total time hang time of a projectile?
The total hang time of a projectile is the time it takes for the projectile to go up and come back down to its original height. It can be calculated using the formula: Total hang time = 2 * time to reach maximum height.
In the oblique projection a projectile reaches the maximum height in 8 seconds it is time of flight is?
If a projectile takes 8 seconds to reach its maximum height, it will take another 8 seconds to return to its original elevation. Presuming it is lauched from flat ground and returns to the ground, its total time in flight would be 16 seconds. If it is launched from a hill, or at a hill, more information would be needed.
What is the initial velocity of a bullet fired verticaly upward from a gun and reaches a height of 7000 ft?
The initial velocity of the bullet can be obtained by using the kinematic equation for projectile motion. Assuming we neglect air resistance, the initial velocity of the bullet fired vertically upward from a gun can be calculated by setting the final velocity as 0 when it reaches the maximum height of 7000 ft. Using the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity, a is the acceleration due to gravity, and s is the total displacement. Solve for u to find the initial velocity of the bullet.
What happens to the total energy of the pendulums as it swings What determines the maximum total energy of the pendulum?
As the pendulum swings, the total energy (kinetic + potential) remains constant if we ignore friction. The maximum total energy of the pendulum is determined by the initial conditions such as the height from which it is released and the velocity. The higher the release point and the greater the initial velocity, the higher the maximum total energy of the pendulum.
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.
Are total length axial length maximum width basal width maximum thickness midsection thickness proximal shoulder angle notch opening and neck width examples of projectile points?
Yes. They are the ATTRIBUTES of the projectile points.
What quantities remain constant in projectile motion?
All that I can think of are: 1.) Gravity 2.) Wind 2.A) wind speed 2.B) direction of wind 3.) Angle of trajectory 4.) Initial speed of projectile 5.) Material through which projectile travels (as in density) 6.) Mass of projectile 7.) Spin 7.A) speed of spin 7.B) axis/axes spining occurs on 8.) Shape of projectile 9.) Temperature of medium projectile is in 10.) Size of projectile (as in height, width, and depth) 11.) Weighting of projectile 12.) Obsturctions to projectile's path In a vaccuum, though, these are the variables: 1.) Speed of object 2.) Obstructions in path 3.) Gravity
How trigonometry is related to to daily activities?
I learned on Khan Academy that for instance 1/2base X height gives you the total displacement or Distance in a vector quantity in comparison to how fast you accelerate and the distance covered. Slope in essence is acceleration. Now for trigonometry, it is figuring out angles, and angles are important in regards to many functions. Lets talk about a cannon's projectile. For instance, i want to figure out a projectiles velocity and displacement from lets say, a 30 degree incline from the flat surface that it is on. If we have the velocity of the the projectile and angle of 30 degrees, we can figure out the distance from where that projectile was launched from, or vice versa.
Given the velocity of a projectile the angle of projection and the mass of the object how do you determine the distance traveled?
Given the initial velocity V, and the angle from the ground A, the total distance travelled X will be: X = 2 V2 cos(A) sin(A) / gwhere "g" is the acceleration due to gravity, on earth g is approximately 9.81 m/s2.You will notice that the mass of the object does not affect the distance traveled. We can derive this by first determining how long the projectile will be in the air. If the initial velocity is V, then the initial vertical velocity is Vsin(A). The vertical velocity will decrease at a rate of 'g' until the vertical velocity reaches zero (known as apogee), and the projectile starts falling down. The time from launch to apogee will be Vsin(A)/g.The time for the projectile to go up is the same as for the projectile to fall down again, so the total time in the airis 2Vsin(A)/g.Assuming we neglect friction, the horizontal velocity is Vcos(A) and does not change. The total distance traveled horizontally is the horizontal speed multiplied by the time spend in the air. So X = 2Vsin(A)/g * Vcos(A) = 2V2cos(A)sin(A)/g.The maximum distance is achived with an angle of 45o. The distance travelled is symmetric around this value, i.e. an angle of 50o will give the same distance as 40o, and an angle of 15owill give the same distance as 75o.
At what angle from a horizontal should a weapon be fired to achieve minimum distance traveled by the projectile?
The weapon should be fired at a 45-degree angle from the horizontal to achieve the minimum distance traveled by the projectile. This angle maximizes the range (horizontal distance) of the projectile by balancing the vertical and horizontal components of its velocity. At any other angle, the total distance traveled would be greater.
Why projectile angle increase when projectile range also increase?
Range of a projectileThe path of this projectile launched from a height y0 has a range d.In physics, a projectile launched with specific initial conditions in a uniform gravity field will have a predictable range. As in Trajectory of a projectile, we will use:g: the gravitational acceleration-usually taken to be 9.80 m/s2 (32 f/s2) near the Earth's surfaceθ: the angle at which the projectile is launchedv: the velocity at which the projectile is launchedy0: the initial height of the projectiled: the total horizontal distance travelled by the projectileWhen neglecting air resistance, the range of a projectile will beIf (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify toso to increase the range θ shoud vary from 0 to 45 and after 45 it starts decreasing.
If you throw a ball and it makes a parabola and all you know is its maximum height and displacement can you find its time in the air?
Yes you can.All you need to know is it's maximum height to find the total time.We are not concerned with displacement because it's horizontal velocity is constant and there are no horizontal forces.Let it's maximum height = hIt's vertical velocity at it's highest position = 0Acceleration due to gravity = gTo find the time it takes the ball to come down from it's highest position can be calculated using the equation, s = ut + (1/2)at2 (Where s= displacement, u= initial velocity, a=acceleration and t=time)=> h = 0 x t + (1/2)gt2=> h = (1/2)gt2=> t2 = 2h/g=> t = √(2h/g)The time it takes for the ball to go up is same as it is to come down because the acceleration is same for both the cases, hence the total time the ball remains in air will be 2t = 2 x √(2h/g)
