Ancient Greek catapults and medical packs
Which calculation would give the vertical component of the launch velocity during launch trial 2?
[value] => Array ( [answer] => 2 [description] =>Reason for the Correct Answer:
The launch velocity is the hypotenuse of a right triangle. Try to use the definitions of sine and cosine to find the components of the launch velocity.
The vertical component will be the opposite side of the 30-degree angle.
Sine theta is defined to be opposite over hypotenuse. So, sin(30°= vy / 40 m/s. Vy = 40 m/s x sin(30°).
) [answers] => Array ( [0] => Array ( [each_answer] =>A. (40m/s) x cos30
) [1] => Array ( [each_answer] =>B. (40m/s) x sin30
) [2] => Array ( [each_answer] =>C. (30m/s) x sin40
) [3] => Array ( [each_answer] =>D. (40m/s) x tan30
) ) ) [1] => Array ( [quiz_unique_key] => 3873426850 [question] =>According to the data above, what effect does increasing launch angle have on launch distance?
[value] => Array ( [answer] => 2 [description] =>Reason for the Correct Answer:
To determine the effect of angle on launch distance, look at trials where only the angles are different but every other variable is the same.
Trials 1, 3, and 5 all use launch speeds of 20m/s. So the only variable that is different is launch angle.
Since for trials 1, 3, and 5 the angle is the only thing that changed and the launch distances increase each time, it can be inferred that increasing angle causes an increase in launch distance.
) [answers] => Array ( [0] => Array ( [each_answer] =>A. Increasing angle has no effect on launch distance.
) [1] => Array ( [each_answer] =>B. Increasing angle causes launch distance to increase.
) [2] => Array ( [each_answer] =>C. Impossible to say since distances both increase and decrease.
) [3] => Array ( [each_answer] =>D. Increasing angle causes launch distance to decrease.
) ) ) [2] => Array ( [quiz_unique_key] => 83407773 [question] =>Out of all the launch trials, what was the largest horizontal component of velocity achieved by the medical packs?
[value] => Array ( [answer] => 4 [description] =>Reason for the Correct Answer:
The medical pack with the largest horizontal component of velocity at launch is not the same as the pack with the greatest launch distance.
The launch velocity is the hypotenuse of a right triangle. Try to use the definitions of sine and cosine to find the components of the launch velocity.
The horizontal component of launch velocity will be the adjacent side of the shot angle. Cosine theta is defined to be adjacent over hypotenuse.
For trial 2 you can solve to find cos(30°) = vₓ /40 m/s. Or, vₓ= 40m/s x cos(30°). This value is greater than the horizontal components of velocity for the other four trials.
) [answers] => Array ( [0] => Array ( [each_answer] =>A. (40m/s) x cos40
) [1] => Array ( [each_answer] =>B. (20m/s) x cos45
) [2] => Array ( [each_answer] =>C. (20m/s) x cos30
) [3] => Array ( [each_answer] =>D. (40m/s) x cos30
) ) ) [3] => Array ( [quiz_unique_key] => 2261298308 [question] =>Compare the total horizontal displacement and vertical displacement of the 5th medical pack launch from the time the 5th pack is shot until the 5th pack hits the ground.
[value] => Array ( [answer] => 3 [description] =>Reason for the Correct Answer:
The pack travels up just as far as it falls down.
Displacement is the difference between the final position and the initial position.
The vertical displacement will be zero since the pack will end up at the same height it started at, so the horizontal displacement will be greater than the vertical displacement.
) [answers] => Array ( [0] => Array ( [each_answer] =>A. The horizontal and vertical displacement are equal since the angle is 45 degrees
) [1] => Array ( [each_answer] =>B. The horizontal displacement was less than the vertical displacement since the medical pack went higher up than it went to the right
) [2] => Array ( [each_answer] =>C. The horizontal displacement is greater than the vertical displacement since the total vertical displacement was zero
) [3] => Array ( [each_answer] =>D. It is impossible to say without knowing the maximum vertical height reached by the medical pack
) ) ) [4] => Array ( [quiz_unique_key] => 2261298308 [question] =>What would be the most reasonable estimate for the launch distance of a medical pack shot at 30 m/s and an angle of 40 degrees?
[value] => Array ( [answer] => 1 [description] =>Reason for the Correct Answer:
Look at launch trials 3 and 4.
The medical pack shot at 20m/s and 40 degrees went 40m. The medical pack shot at 40m/s and 40 degrees went 161m.
The launch distance of the 30m/s medical pack should be between the 20m/s and 40m/s medical packs. Therefore, eliminate answer choice 39m.
From the data, it is apparent that changing the angle of projection significantly changes the distance. Both 50m and 158m are too close to 40m and 161m to be reasonable choices. The correct answer is 90m.
) [answers] => Array ( [0] => Array ( [each_answer] =>A. 90 meters
) [1] => Array ( [each_answer] =>B. 50 meters
) [2] => Array ( [each_answer] =>C. 39 meters
) [3] => Array ( [each_answer] =>D. 158 meters
) ) ) ) [total_question] => 5 [correct_answers] => Array ( [556515|1] => B [556515|2] => B [556515|3] => D [556515|4] => C [556515|5] => A ) [hide_display_feedback_settings] => [hide_solutions] => )