PHYS207 Navros 2D Motion Homework Assignment

PHYS207 Navros 2D Motion Homework Assignment

Problem 1: Projectile Launched at an Angle

1. Scenario: A projectile is launched on level ground at an angle $heta$ above the horizontal with initial speed $v_i$. Air resistance is ignored.
      - a) Calculation of Elapsed Time ($riangle t$): The time that the projectile is in the air can be calculated using the formula:
        - $riangle t = \frac{2 v_i imes ext{sin}( heta)}{g}$
        where $g$ is the acceleration due to gravity.
      - b) Total Horizontal Displacement ($riangle x$): The horizontal displacement of the projectile can be calculated as:
        - $riangle x = v_i imes ext{cos}( heta) imes riangle t$
        Substituting the expression for elapsed time,
        - $riangle x = v_i imes ext{cos}( heta) imes \frac{2 v_i imes ext{sin}( heta)}{g} = \frac{2 v_i^2 imes ext{sin}( heta) imes ext{cos}( heta)}{g}$
        Using the identity $ext{sin}(2 heta) = 2 imes ext{sin}( heta) imes ext{cos}( heta)$, we further simplify:
        - $riangle x = \frac{v_i^2 imes ext{sin}(2 heta)}{g}$

Problem 2: Football Thrown by a Quarterback

1. Scenario: A quarterback throws a football with components of initial velocity, namely an upward velocity component of $v_y$ and a horizontal velocity component of $v_x$. Air resistance is ignored.
      - a) Time to Reach Maximum Height: The time to reach the highest point of the trajectory can be determined by:
        - $t = \frac{v_y}{g}$
      - b) Height at Maximum Point: The height at the highest point is given by:
        - $h = v_y imes t - \frac{1}{2} g t^2$
        Replacing $t$ with $\frac{v_y}{g}$ gives:
        - $h = v_y imes \frac{v_y}{g} - \frac{1}{2} g imes \frac{v_y^2}{g^2} = \frac{v_y^2}{2g}$
      - c) Time for Return to Original Level: The time for the football to return to the original launch level is twice the time to reach maximum height:
        - $t_{ ext{return}} = 2t = \frac{2v_y}{g}$
      - Comparison: This calculated time relates to part (a) as it is twice as long since it includes both ascent and descent time.
      - d) Horizontal Distance Travelled: The horizontal distance during the flight can be assessed as:
        - $riangle x = v_x imes t_{ ext{return}} = v_x imes \frac{2v_y}{g}$
      - e) Graphs: Draw the following graphs for the motion:
        - x-t Graph: Shows linear increase in horizontal displacement over time.
        - y-t Graph: Parabolic shape showing rise to maximum height and return to original level.
        - vx-t Graph: Constant horizontal velocity over time (no air resistance).
        - vy-t Graph: Linear decrease to zero at the max height followed by linear increase back to original level.

Problem 3: Projectile Avoiding Temperature Inversion Layer

1. Scenario: A projectile is launched from ground level with no air resistance, needing to avoid penetrating a temperature inversion layer at height $h$.
      - a) Maximum Launch Speed: The maximum launch speed if shot straight up can be found using:
        - $v_{max} = ext{sqrt}(2gh)$. This is derived from the energy conservation principle where potential energy is equal to kinetic energy.
      - b) Maximum Angle for Twice the Speed: If the launch speed is twice the maximum calculated in part (a):
        - $v = 2v_{max} = 2 ext{sqrt}(2gh)$
        The launch angle that maximizes horizontal range can be determined to be $45^ ext{o}$ above the horizontal, since this provides optimal range for any velocity in projectile motion.
      - c) Horizontal Distance Landed ($riangle x$): As the range formula for launched at an angle $heta$ becomes:
        - $riangle x = \frac{v^2 imes ext{sin}(2 heta)}{g}$
        Using $heta = 45^ ext{o}$ gives:
        - $riangle x = \frac{(2v_{max})^2}{g} = \frac{4(2gh)}{g} = 8h$

Problem 4: Battleship Firing Shells

1. Scenario: A battleship fires two shells simultaneously with the same initial speed $v_i$ at different angles. The first is aimed at a steep angle to hit target A, and the second at a shallow angle for target B.
      - a) Time to Reach Maximum Height: For a shell fired at angle $heta$, the time to reach maximum height is:
        - $t_{max} = \frac{v_i imes ext{sin}( heta)}{g}$
      - b) Time to Reach Target: The time to reach the target can be derived as follows considering the full flight path:
        - $t_{target} = \frac{2v_i imes ext{sin}( heta)}{g}$
      - c) Which Target is Struck First: Compare $t_{max}$ and $t_{target}$ to determine that target A, which is aimed at a steep angle, has less time to reach maximum height than target B which would have a longer horizontal distance and time due to the shallower angle, hence target A is struck first.