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A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height?
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A projectile is projected with three eye cap 4 j cap 5K cap metre ...
Projectile Motion in 3D Space
Projectile motion is a type of motion in which an object is thrown into the air and moves along a parabolic path under the influence of gravity. In this question, we are given the velocity vector of the projectile at the point of projection and we need to find its displacement when it hits the xy plane and its location when it is at its maximum height. Let's solve each part of the question separately.
Displacement when it hits xy plane
We are given the velocity vector of the projectile as v = 3i + 4j + 5k m/s. The initial position of the projectile is taken as the origin. Since the projectile is moving in a three-dimensional space, we need to break down the velocity vector into its x, y, and z components.
vx = 3 m/s, vy = 4 m/s, vz = 5 m/s
We can use the kinematic equation for displacement in the z-direction to find the time taken by the projectile to hit the xy plane.
z = vz*t - 1/2*g*t^2
where z is the displacement in the z-direction, vz is the initial velocity in the z-direction, g is the acceleration due to gravity, and t is the time taken.
At the point of impact, z = 0. Solving the above equation for t, we get:
t = 1 s
Substituting this value of t in the x and y equations of motion, we can find the displacement of the projectile in the x and y directions.
x = vx*t = 3 m
y = vy*t = 4 m
Therefore, the displacement of the projectile when it hits the xy plane is (3i + 4j) m.
Location of Particle at Maximum Height
To find the location of the particle at its maximum height, we need to first find the time taken to reach the maximum height. The projectile will reach its maximum height when its vertical velocity becomes zero. At this point, the projectile will start to fall back to the ground.
Let's find the time taken to reach the maximum height.
vz = uz - g*t
where vz is the final velocity in the z-direction, uz is the initial velocity in the z-direction, g is the acceleration due to gravity, and t is the time taken.
At the maximum height, vz = 0 and uz = 5 m/s. Solving the above equation for t, we get:
t = 0.5 s
Substituting this value of t in the z equation of motion, we can find the maximum height reached by the projectile.
z = uz*t - 1/2*g*t^2
z = 5*0.5 - 1/2*10*(0.5)^2
z = 1.25 m
Therefore, the maximum height reached by the projectile is 1.25 m.
To find the location of the particle at its maximum height, we can use the x and y equations of motion.
x = vx*t = 1.5 m
y = vy*t = 2 m
z = 1.25 m
Therefore, the location of the particle when it is at its maximum height is (1.5i + 2j + 1.25k) m.
Projectile motion is a type of motion in which an object is thrown into the air and moves along a parabolic path under the influence of gravity. In this question, we are given the velocity vector of the projectile at the point of projection and we need to find its displacement when it hits the xy plane and its location when it is at its maximum height. Let's solve each part of the question separately.
Displacement when it hits xy plane
We are given the velocity vector of the projectile as v = 3i + 4j + 5k m/s. The initial position of the projectile is taken as the origin. Since the projectile is moving in a three-dimensional space, we need to break down the velocity vector into its x, y, and z components.
vx = 3 m/s, vy = 4 m/s, vz = 5 m/s
We can use the kinematic equation for displacement in the z-direction to find the time taken by the projectile to hit the xy plane.
z = vz*t - 1/2*g*t^2
where z is the displacement in the z-direction, vz is the initial velocity in the z-direction, g is the acceleration due to gravity, and t is the time taken.
At the point of impact, z = 0. Solving the above equation for t, we get:
t = 1 s
Substituting this value of t in the x and y equations of motion, we can find the displacement of the projectile in the x and y directions.
x = vx*t = 3 m
y = vy*t = 4 m
Therefore, the displacement of the projectile when it hits the xy plane is (3i + 4j) m.
Location of Particle at Maximum Height
To find the location of the particle at its maximum height, we need to first find the time taken to reach the maximum height. The projectile will reach its maximum height when its vertical velocity becomes zero. At this point, the projectile will start to fall back to the ground.
Let's find the time taken to reach the maximum height.
vz = uz - g*t
where vz is the final velocity in the z-direction, uz is the initial velocity in the z-direction, g is the acceleration due to gravity, and t is the time taken.
At the maximum height, vz = 0 and uz = 5 m/s. Solving the above equation for t, we get:
t = 0.5 s
Substituting this value of t in the z equation of motion, we can find the maximum height reached by the projectile.
z = uz*t - 1/2*g*t^2
z = 5*0.5 - 1/2*10*(0.5)^2
z = 1.25 m
Therefore, the maximum height reached by the projectile is 1.25 m.
To find the location of the particle at its maximum height, we can use the x and y equations of motion.
x = vx*t = 1.5 m
y = vy*t = 2 m
z = 1.25 m
Therefore, the location of the particle when it is at its maximum height is (1.5i + 2j + 1.25k) m.
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A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height?
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A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height?.
A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height?.
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A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height?, a detailed solution for A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height? has been provided alongside types of A projectile is projected with three eye cap 4 j cap 5K cap metre per second point of projection is taken as origin and is not vertically upward directions are taken as positive xyz directions respectively take G equals to 10 m/s all the units are in SI system question number 1 it's displacement when it hits xy plane Question number 2 location of particle when it is at its maximum height? theory, EduRev gives you an
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