Projectile Motion: Position and Velocity Calculation

What is the position and velocity of a ball kicked into the air at an angle of 30° above the horizontal with an initial velocity of 20 m/s after 1.5 seconds? The question deals with the physics concept of projectile motion. By applying relevant formulae for position and velocity in horizontal and vertical directions, we can derive the position and velocity of the ball after 1.5 seconds.

Projectile motion is a fascinating concept in physics that involves the motion of objects in two dimensions under the influence of gravity. When a ball is kicked or thrown into the air at an angle, we can analyze its position and velocity at different time intervals to understand its trajectory.

Position Calculation:

For the horizontal direction, we use the equation: X = Vx*t, where Vx = V * cos(θ). Substituting the values, we get X = 20 * cos(30) * 1.5.

For the vertical direction, we use the equation: Y = Vy*t - 0.5gt^2, where Vy = V * sin(θ) and g is the acceleration due to gravity. Substituting the values, we get Y = 20 * sin(30) * 1.5 - 0.5 * 9.8 * 1.5^2.

Velocity Calculation:

For the horizontal component of velocity, there is no acceleration, so Vx remains constant at 20 * cos(30) m/s.

For the vertical component of velocity at 1.5 seconds, we calculate Vy(t) = Vy - g * t. Substituting the values, we get Vy(1.5) = 20 * sin(30) - 9.8 * 1.5 m/s.

Therefore, the position of the ball after 1.5 seconds is X = 17.32 m and Y = 8.78 m, while the velocity components are Vx = 17.32 m/s and Vy(1.5) = 5.9 m/s.
← Calculating kinetic energy of a bike and rider How to calculate speed and direction after collision of two billiard balls →

More Posts: