Distinguish between a vector and scalar - Leaving Cert Physics - Question 14(a) - 2022

To draw the arrangement:

• Use a horizontal line to represent the ground.
• Draw two arrows originating from a common point; one arrow for each vector, labeled with their respective magnitudes and directions.
• Indicate the resultant vector with an arrow connecting the heads of the two vectors, and label it 'Resultant'.

The initial velocity is given as 150 m s⁻¹ at an angle of 20°:

• Horizontal component, $v_x = v imes ext{cos}( heta)$:

  $v_x = 150 imes ext{cos}(20°) ≈ 141.4 ext{ m s}^{-1}$

• Vertical component, $v_y = v imes ext{sin}( heta)$:

  $v_y = 150 imes ext{sin}(20°) ≈ 51.3 ext{ m s}^{-1}$

Using the vertical motion equation:

$v_y = u_y - gt$

Where:

• $u_y = 51.3 ext{ m s}^{-1}$
• $g = 9.8 ext{ m s}^{-2}$
• $t = 8 s$

Substituting values:

$v_y = 51.3 - (9.8 imes 8) = 51.3 - 78.4 = -27.1 ext{ m s}^{-1}$

The negative sign indicates downward direction.

Next, calculate the final horizontal velocity, which remains unchanged:

$v_x = 141.4 ext{ m s}^{-1}$

Finally, the magnitude of the resultant velocity can be calculated using Pythagoras theorem:

$|v| = ext{sqrt}(v_x^2 + v_y^2) = ext{sqrt}(141.4^2 + (-27.1)^2) ≈ 143.5 ext{ m s}^{-1}$

The direction can be calculated using:

heta = ext{tan}^{-1}igg(rac{v_y}{v_x}igg) ≈ ext{tan}^{-1}igg(rac{-27.1}{141.4}igg) ≈ -10.9°

Hence, the final velocity is approximately 143.5 m s⁻¹ at an angle of 10.9° below the horizontal.