(Solved): PHYS 125L. Lab. #S1 Proiectile Motion with and Without Air Resistance Objectives: To determine the r ...
PHYS 125L. Lab. #S1 Proiectile Motion with and Without Air Resistance Objectives: To determine the relationship between the angle of projection from a golf ball is launched, the initial velocity of the ball, the vertical and horizontal positions of the ball. To determine the effects of air resistance (air drag) on the trajectory of the golf ball and the dependence of the air resistance on the mass and diameter of the ball. Warning: Students are not allowed to share answers or results. Any simulation lab report that is deemed to be copied from another student or plagiarized will be assigned zero grade. All entries in the tables have to be type-written. All answers have to also be typewritten. (No hand-written work will be accepted). Procedure: Open Projectile Motion Lab using this link: hups://phet.colorado.sdu/sims/html/projectile-motion/latest projectile-motion_en.html and click the play button. Once the simulation opens, chose "Lab" option. Fill out the table on the next page and carry out the required calculations with the equations that are provided. Please answer the questions that follow your calculations thoroughly. Part 1: Projectile Motion Without Air Resistance Make sure that the "Air Resistance" box is unchecked and that acceleration due to gravity is set at
g=9.81(m)/(s^(2)). Here are the entries in the table: Initial velocity,
v_(0)((m)/(s))Projection angle,
\theta \deg . Simulated vertical position (height),
y, (m) Calculated vertical position (height),
y_(,)(m) Simulated horizontal position (range),
x_(,)(m)Calculated horizontal position (range),
x_(c)(m) Time,
t(s).\table[[
v_(o ),
\theta \deg ,
t,
y_(, ),
^(**)y_(c),% Difference,
x,,
^(****)x_(c),% Difference],[,,,,,,,,],[14,30,0.60,,,,,,],[14,45,0.60,,,,,,],[14,60,0.60,,,,,,],[18,30,0.80,,,,,,],[18,30,1.40,,,,,,],[22,30,0.80,,,,,,],[22,45,1.59,,,,,,],[22,60,3.88,,,,,,],[18,30,1.83,,,,,,],[18,60,3.18,,,,,,]] Calculated vertical position (height),
y_(c)=v_(c)sin\theta t-(1)/(2)gt^(2)** Calculated horizontal position (range),
x_(e)=v_(e)cos\theta tQuestions: Using
v_(e)=14(m)/(s)and
t=0.60s, which angle gives the largest
y_(e)? Explain using trigonometric argument. Using
v_(e)=14(m)/(s)and
t=0.60s, which angle gives the largest
x_(e)? Explain using trigonometric argument. Using
v_(e)=22(m)/(s),\theta =45\deg , and
t=1.59s, calculate the maximum vertical position (maximum height),
y_(max)=(1)/(2)gt^(2). Does this match the
y, in the table? What is the vertical speed at the maximum height? Using
v_(e)=22(m)/(s),\theta =60\deg , and
t=3.88s, calculate the maximum horizontal position (maximum range),
x_(max)=(v_(e)^(2)sin(2\theta ))/(g). Does this match the
x, in the table? Using
v_(e)=18(m)/(s),\theta =30\deg , and
t=1.83s, calculate the maximum horizontal position (maximum range),
x_(max)=(v_(e)^(2)sin(2\theta ))/(g). Do the same calculation for
v_(e)-18(m)/(s),\theta =60\deg , and
t=3.18s. Why are the results the same? Hint: Think about the term
sin(2\theta ).
