the escalator will never have an inclination of 45deg the standard of angle i 30and 35 deg only
The initial magnitude of the velocity is sqrt(5) times the horizontal component. This results in a velocity vector that is inclined at an angle of arctan(2) ≈ 63.43 degrees with respect to the horizontal.
The vertical velocity component of the ball can be found by multiplying the initial speed (31 m/s) by the sine of the launch angle (35 degrees). Vertical velocity = 31 m/s * sin(35) ≈ 17.7 m/s. The vertical velocity component is approximately 17.7 m/s.
The angle of projection affects the maximum height by determining the vertical and horizontal components of the initial velocity. At 90 degrees (vertical), all the initial velocity is vertical which results in maximum height. As the angle decreases from 90 degrees, the vertical component decreases, leading to a lower maximum height.
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When an object is launched at a 45-degree angle, it splits the initial velocity into horizontal and vertical components equally. This allows for the maximum range because the horizontal component remains constant throughout the flight, while the vertical component decreases due to gravity. This balance between horizontal and vertical components at 45 degrees results in the maximum distance traveled before hitting the ground.
The initial magnitude of the velocity is sqrt(5) times the horizontal component. This results in a velocity vector that is inclined at an angle of arctan(2) ≈ 63.43 degrees with respect to the horizontal.
The vertical velocity component of the ball can be found by multiplying the initial speed (31 m/s) by the sine of the launch angle (35 degrees). Vertical velocity = 31 m/s * sin(35) ≈ 17.7 m/s. The vertical velocity component is approximately 17.7 m/s.
The angle of projection affects the maximum height by determining the vertical and horizontal components of the initial velocity. At 90 degrees (vertical), all the initial velocity is vertical which results in maximum height. As the angle decreases from 90 degrees, the vertical component decreases, leading to a lower maximum height.
Multiply the speed by the cosine of the angle (25 degrees in this case). For the vertical velocity, multiply by the sine of 25 degrees.Multiply the speed by the cosine of the angle (25 degrees in this case). For the vertical velocity, multiply by the sine of 25 degrees.Multiply the speed by the cosine of the angle (25 degrees in this case). For the vertical velocity, multiply by the sine of 25 degrees.Multiply the speed by the cosine of the angle (25 degrees in this case). For the vertical velocity, multiply by the sine of 25 degrees.
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When an object is launched at a 45-degree angle, it splits the initial velocity into horizontal and vertical components equally. This allows for the maximum range because the horizontal component remains constant throughout the flight, while the vertical component decreases due to gravity. This balance between horizontal and vertical components at 45 degrees results in the maximum distance traveled before hitting the ground.
The vertical velocity at the top of the path of a projectile thrown straight up is 0 m/s because it momentarily stops before falling back down. For a projectile launched at an angle, the vertical velocity at the top of the path depends on the initial velocity and launch angle, but it will also momentarily be 0 m/s before changing direction.
You have a right triangle and can use trig. Degree mode. tan(theta) = adjacent/opposite( y component ) tan( 60 degrees) = (5 m/s)/(y comp.) y component = 5 m/s)/(tan 60 degrees) = 2.887 m/s ( you can call it 3 m/s ) -----------------------------------------------
The initial velocity of the ball can be calculated using the kinematic equation for projectile motion. By using the vertical component of velocity (V0y) and the time of flight, we can determine the initial velocity needed for the ball to reach the hoop. The velocity components are V0x = V0 * cos(θ) and V0y = V0 * sin(θ), where θ is the initial angle. The time of flight in this case is determined by the vertical motion of the ball, and it can be found by using the equation of motion for the vertical direction, considering the initial vertical velocity, the gravitational acceleration, and the vertical displacement of the ball. Once these values are calculated, the initial velocity can be computed by combining the horizontal and vertical components of the motion.
If you ignore the effect of the air grabbing at it and only figure in gravity, then the horizontal component of velocity is constant, from the time the stone leaves your hand until the time it hits the ground. Makes no difference whether you toss it up, down, horizontal, or on a slant. Also makes no difference whether it's a cannonball, a stone, or a bullet.
Can't say. It depends on the release velocity (muzzle velocity).The maximum horizontal distance always results from an angle of 45 degrees, regardless of the release velocity.
The horizontal velocity component of the ball can be calculated using the formula: horizontal velocity = initial velocity * cos(angle). Substituting the values, we get: horizontal velocity = 31 m/s * cos(35 degrees) ≈ 25.3 m/s.