Vector calculations on velocity

• What is the reason for the cannonball's path being curved?

  The vertical component of the velocity increases with time while the horizontal component is constant.
• How can the magnitude of the resultant velocity of a projectile be calculated?

  It can be calculated from the components \( v_x \) and \( v_y \) using Pythagoras's theorem.
• What is the formula used to calculate the resultant velocity?

  The resultant velocity \( v \) is calculated using \( v = \sqrt{v_x^2 + v_y^2} \).
• What is the initial horizontal velocity of the water from the hose pipe?

  The initial horizontal velocity is \( 7.0 \, \text{m s}^{-1} \).
• Why does the horizontal component of the velocity remain constant at \( 7.0 \, \text{m s}^{-1} \)?

  The horizontal component remains constant because the velocity is traveling at a constant speed.
• How can you show that the water takes about \( 0.5 \, \text{s} \) to travel from the end of the pipe to the ground?

  You can show this by calculating the time using the horizontal distance and velocity: \( t = \frac{d}{v} = \frac{3.5 \, \text{m}}{7.0 \, \text{m s}^{-1}} \approx 0.5 \, \text{s} \).
• What is the relationship between horizontal distance, horizontal velocity, and time for the water from the hose pipe?

  • Horizontal distance \( d = 3.5 \, \text{m} \)
  • Horizontal velocity \( v = 7.0 \, \text{m s}^{-1} \)
  • Time \( t \) can be calculated as \( t = \frac{d}{v} \approx 0.5 \, \text{s} \)
• What are the components of velocity for a projectile like the cannonball?

  • Horizontal component \( v_x \) remains constant
  • Vertical component \( v_y \) increases with time
  • Resultant velocity \( v \) is calculated using \( v = \sqrt{v_x^2 + v_y^2} \)