d=1/2at^2
80=1/2(9.80)t^2
80=4.90t^2
t^2=80/4.90
t=square root 16.33
t=4.04s
v=d/t
v=1330/4.04
v=330 m/s
If the initial velocity is v, at an angle x to the horizontal, then the vertical component is v*sin(x) and the horizontal component is v*cos(x).
There are several definitions. not just one. Average velocity in a direction = Average displacement (distance) in that direction/time Instantaneous velocity in a direction = derivative of displacement in that direction with respect to time Average velocity in a direction = Initial velocity in that direction + Average acceleration in that direction * time Instantaneous velocity in a direction = Definite integral of acceleration in that direction with respect to time, with initial velocity at t = 0 Then there are others in which time is eliminated.
Positive acceleration.
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
AccelerationWhen the velocity of an object increases or decreases, that means it has accelerated. Acceleration is defined as the rate of change of velocity.If an object's final velocity is greater than its initial velocity, that indicates positive acceleration. If an object's final velocity is less than its initial velocity -- if, say, it slows down and comes to a stop -- then that indicates negative acceleration. Deceleration is another way of saying negative acceleration. But . . .It is good idea to avoid using the term deceleration, because an object that is experiencing negative acceleration may slow down, come to a stop momentarily, and then reverse direction and speed up -- IN THE OPPOSITE DIRECTION!You can think of it this way: When an object is slowing down, its acceleration is in the direction opposite to its motion. We think of that as negative acceleration.
The horizontal component of the initial velocity of the ball is the velocity in the horizontal direction at the moment the ball is launched. It represents the speed and direction at which the ball is moving side-to-side.
The initial direction of a projectile's velocity is typically determined by the angle at which it is launched relative to the horizontal plane. This angle will influence both the horizontal and vertical components of the velocity.
Yes, in projectile motion, the vertical component of motion is influenced by the initial velocity in the vertical direction. The horizontal and vertical components of motion are independent of each other, with the horizontal component being influenced by the initial velocity in the horizontal direction.
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.
In projectile motion, since , there's no force in the horizontal direction which can change the horizontal motion therefore the horizotal velocity remains conserved Vx=Vox= Vocos theta by using above formula , constant horizontal initial or final velocity can be found. since Initial = final horizontal velocity.
To find the horizontal displacement of the ball, you can use the equation of motion in the horizontal direction, which is given by: horizontal displacement = initial velocity * time * cos(angle). Given the initial velocity is 25.0 m/s and the angle is 35 degrees, the horizontal displacement can be calculated once the time of flight is known.
The horizontal speed of the projectile remains constant as there is no force acting in the horizontal direction to change it. Therefore, the horizontal speed of the projectile after 3 seconds will remain at 4 m/s.
In projectile motion, the only force acting horizontally is the initial velocity, which does not change over time in the absence of external horizontal forces. This means that the acceleration in the horizontal direction is constant and therefore zero because there are no forces causing a change in velocity in that direction.
If the initial velocity is v, at an angle x to the horizontal, then the vertical component is v*sin(x) and the horizontal component is v*cos(x).
The curved path is called a parabolic trajectory. This occurs due to the combination of the projectile's initial horizontal velocity and the acceleration due to gravity in the vertical direction.
Using the projectile motion equations and given the initial velocity and angle, we can calculate the time the shell is in the air. Then, we can find the horizontal range by multiplying the time of flight by the horizontal component of the initial velocity. The horizontal range in this case is about 1056 meters.
When the initial velocity is zero, the object is at rest, and there is no motion in the direction of the velocity. This means that the object will not have any kinetic energy due to its velocity at the initial moment.