The vertical velocity is 0. The horizontal velocity is constant during the entire trajectory (and may be zero).
To achieve nonzero speed at the top of the trajectory, you should throw the projectile upward with an initial velocity greater than zero. This will allow the projectile to continue moving upward even at the top of its trajectory before it begins to fall back down due to gravity.
No, assuming no air resistance, there will be a constant downward acceleration of 9.8 meters per second square (assuming standard gravity). The vertical component of the velocity will be zero at the top of the trajectory.
Assuming that there is no velocity in the horizontal direction, then the velocity at that instant is zero.
Acceleration at the point of zero vertical velocity will be equivalent to gravitational acceleration on that body. On Earth, for example, this is around 9.8 meters per second per second (9.8m/s2).
At the top of the trajectory, when the ball momentarily stops before falling back down, its kinetic energy is 0 J (as it stops moving) and its potential energy is equal to the initial potential energy of 100 J. So, the total energy at the top of the trajectory is 100 J.
A projectile has minimum speed at the top of the trajectory.
To achieve nonzero speed at the top of the trajectory, you should throw the projectile upward with an initial velocity greater than zero. This will allow the projectile to continue moving upward even at the top of its trajectory before it begins to fall back down due to gravity.
No, assuming no air resistance, there will be a constant downward acceleration of 9.8 meters per second square (assuming standard gravity). The vertical component of the velocity will be zero at the top of the trajectory.
Assuming that there is no velocity in the horizontal direction, then the velocity at that instant is zero.
Acceleration at the point of zero vertical velocity will be equivalent to gravitational acceleration on that body. On Earth, for example, this is around 9.8 meters per second per second (9.8m/s2).
At the top of the trajectory, when the ball momentarily stops before falling back down, its kinetic energy is 0 J (as it stops moving) and its potential energy is equal to the initial potential energy of 100 J. So, the total energy at the top of the trajectory is 100 J.
Without air friction, the horizontal component of the velocity will be constant. The vertical component of the velocity will be a maximum at the lowest point in its motion and at a minimum at the highest point in its motion. Therefore the minimum is at the highest point in its motion- Potential energy max Kinetic Energy min and the maximum is at its lowest point in the motion- KE is max PE min
The projectile have minimum speed when it is in top of prabolic and it have max sped when it is in intial point
The vertical speed of a projectile at the top of its flight is zero, as it momentarily comes to a stop before beginning its descent due to gravity.
The acceleration of a ball at the top of its trajectory when thrown straight upward is equal to the acceleration due to gravity, which is approximately 9.81 m/s^2 downward. At the top of its trajectory, the ball momentarily comes to a stop before reversing direction and accelerating downward.
a roll shot uses top-spin and an arching trajectory
Ballistic spin drift is the tendency of a spinning projectile (such as a bullet) to drift slightly to the side due to the gyroscopic effect created by its spin. This effect influences the trajectory of the projectile and can cause it to deviate from its intended path at longer ranges. Shooters need to account for ballistic spin drift when calculating their shots for increased accuracy.