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Which is the total velocity of a projectile at maximum height?
TabishAhmed
Wiki User
∙ 13y agoFind the velocity in the vertical direction in meters per second. Enter it into the equation 4.9t^2 + (the velocity)t + (initial height). Find where this function is at a max. This is easiest when you use a graphing calculator.
Wiki User
∙ 12y agoThroughout the flight, there is a conversion between gravitational potential energy (which goes up with height) and kinetic energy (which determines speed). So if air resistance is tiny enough to be neglected, the maximum speed would be when the projectile is highest and the minimum speed would be when the projectile is lowest.
Wiki User
∙ 10y agoIf the projectile was traveling straight no with zero horizontal velocity,
then its speed would be zero at its maximum height.
If the projectile had non-zero horizontal velocity during its flight (such as
a thrown baseball), its speed would not be zero at its maximum height.
In that case, its speed at is maximum height would be equal to its horizontal
speed during its flight (assuming that its horizontal speed remains constant).
It's the vertical component of velocity that's zero at the top of the arc.
Wiki User
∙ 9y agoThe a projectile that goes vertically up reaches its maximum height, its speed will be zero. In the more general case, the vertical component of its velocity will be zero; however, the projectile may also have a horizontal component. In summary, assuming that air resistance is negligible:
* The horizontal component of velocity doesn't change over time. At the highest point, it will be the same to the original horizontal component of velocity.
* The vertical component of velocity will be zero at the highest point.
Wiki User
∙ 13y agoThe same as anywhere else (assuming there is no air resistance). Near Earth's surface, that would be about 9.8 meters per second squared.
Wiki User
∙ 12y agoyes
Duane Anymouse
Wiki User
∙ 12y agoThe apex.
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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 are the calculations for launching a projectile at an angle for distance i.e. a shell shot from a tank?
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.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.
What is the initial velocity that must be given to a 1kg mass if it is to move to a height of 20m. what is the total energy?
After 20 m all kinetic energy have been transformed to potential energy.Epot=m*g*h=0.15*9.81*20=29.4 Joule
Seepage velocity and discharge velocity?
Discharge Velocity is obtained by Dividing the Total Discharge by the total cross Sectional Area , Where Total cross sectional area Consists of void+solid. In contrast .. Seepage Velocity is defined as the total discharge by the Area of voids only. So Seepage velocity always greater than Discharge Velocity.
How is average velocity of a body calculated when its velocity changes at a non-uniform rate?
average velocity=displacement/total time taken
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.
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.
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.
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 are the calculations for launching a projectile at an angle for distance i.e. a shell shot from a tank?
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.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.
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.
What is the total energy of a projectile if its PE at the apex is 1050j and that value is twice half the maximum kinetic energy of the projectile at landing?
This is a trick question with the 'twice half' words. The energy is 1050 J which is PE at the apex, it is travelling vertically, there is no air resistance, and that energy has been totally converted to KE when the projectile hits the ground.
Total distance divided by total time?
total velocity * * * * * It is the average speed, not velocity which is a vector.
How do you find the initial velocity of a projectile given the height of the launch site and distance from the base of the launch site to the projectile's landing spot?
The way to understand projectiles is to treat their speeds as two different speeds. Treat one speed as how fast the projectile moves up or down relative to the ground, and treat the other speed as how fast the object moves left to right. This is called splitting the velocity into components. Each component is independent of the other, meaning that no matter what happens to the x (the horizontal speed) component, the y (the vertical component) will be unaffected. To determine these components, you will need to know how long the object was in the air. Take the total x component and divide it by the time the object is in the air. That is the x component. Then take that and multiply it by the inverse cosine function of whatever angle the projectile was launched at. That will give you the inital velocity of the object. --An AP Physics Student Bored in Study Hall
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)
What is the initial velocity that must be given to a 1kg mass if it is to move to a height of 20m. what is the total energy?
After 20 m all kinetic energy have been transformed to potential energy.Epot=m*g*h=0.15*9.81*20=29.4 Joule
