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Get the value of initial velocity. Get the angle of projection. Break initial velocity into components along x and y axis. Apply the equation of motion .

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Q: How to find the height of the projectile launch if the angle velocity and the distance are given?
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Is initial velocity y the same for an angled launch and a horizontal launch if you use the same projectile launcher?

Yes. They will both initially be moving at the same speed.


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.


If the launch angle is 15 degrees and the initial velocity is 50.0 meters per second what is the range?

If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.


How does angle affect the distance of a projectile?

yes it does. you see if you have it set up at a a 90 degree angle it will go further than it would of a 10 degree angle A projectile leaving the ground at an angle of 45 degrees will attain the maximum range. Fire it straight up and it will fall back to its launch location (wind effects etc. ignored). Fire it horizontally and it will hit the ground very much the same time as if it was dropped from its launch platform at the same time. That would not be very far.


How do you find the maximum range and height for angles 30 45 and 60?

1. You need to know the velocity of the projectile (V0) 2. The expressions for the range and height assume no air resistance (in vacuum) 3. The units must be consistent e.g. metres and g = 9.81 m/s2 Range in metres for 30 degree launch angle = sin 60 x V02 / 9.81 Range in metres for 45 degree launch angle = sin 90 x V02 / 9.81 Range in metres for 60 degree launch angle = sin 120 x V02 / 9.81 Max. height in metres for 30 degree launch angle = (V0 x sin 30)2 / 2g Max. height in metres for 45 degree launch angle = (V0 x sin 45)2 / 2g Max. height in metres for 60 degree launch angle = (V0 x sin 60)2 / 2g 2g is of course 9.81 x 2 = 19.62 m/s2 For interest, at 45 degree launch angle the max. height is 25% of the range.

Related questions

What do you you need to know to determine how far a projectile travels horizontally?

initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)


What do you need to know to determine how far a projectile travels horizontally?

initial velocity, angle of launch, height above ground When a projectile is launched you can calculate how far it travels horizontally if you know the height above ground it was launched from, initial velocity and the angle it was launched at. 1) Determine how long it will be in the air based on how far it has to fall (this is why you need the height above ground). 2) Use your initial velocity to determine the horizontal component of velocity 3) distance travelled horizontally = time in air (part 1) x horizontal velocity (part 2)


Is initial velocity y the same for an angled launch and a horizontal launch if you use the same projectile launcher?

Yes. They will both initially be moving at the same speed.


What makes up a projectile motion?

Some of the factors that determine the movements of a projectile include: air resistance, force of gravity, initial launch velocity, the angle a projectile was launched at, and the objects initial elevation.


What is the horizontal acceleration of a projectile as its position changes?

In the usual simple treatment of projectile motion, the horizontal component of the projectile's velocity is assumed to be constant, and is equal to the magnitude of the initial (launch) velocity multiplied by the cosine of the elevation angle at the time of launch.


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 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 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


What is the equation to calculate the maximum height of an object given an initial velocity and initial launch height?

height=acceletation(t^2) + velocity(t) + initial height take (T final - T initial) /2 and place it in for time and there you go


If the launch angle is 15 degrees and the initial velocity is 50.0 meters per second what is the range?

If the initial velocity is 50 meters per second and the launch angle is 15 degrees what is the maximum height? Explain.


What shape does a projectile make?

In a perfect system, with no air resistance, the arc that a projectile moves through is a parabola. The shape of the parabola is dependent of various parameters including the initial velocity (speed and angle of launch) as well as the prevailing gravity. It could also describe a circle if the launch criteria are just right for the gravity, such as a satellite orbiting the Earth.


What should be the initial velocity of a rocket if it to hit a target 1000 km away?

initial velocity would be ZERO before launch. To calculate the velocity you would need to hit that target at that distance you would need to know the mass of the rocket and the angle of launch or trajectory simplifying it