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Sam has made this table for his experiment and he draw that graph below. - math calculation: information table (Data): Lunching angle 0 () (5.675-5.015) Range Trial 1 Range Trial 2 Average Range (m) * slope (trial 2) = = 6.6 (0.8660-0.7660) 20 Sin(20) s3 (m) s3 (m) 25 50 0.7660 5.03 5.00 5.015 Vosin20 To find Vg let's take any value for (Range value in the trial ) from this equation s, = 30 60 0.8660 5.66 5.69 5.675 35 70 0.9396 6.19 6.15 6.17 v.(0. 7660) 0.98480 6.435 40 80 6.44 6.43 1) 5.015 = 45 90 1 6.64 6.69 6.665 Vosin20 from this equation S, = , we can put that sin20 as a variable on the X- axis, Vo = 8.0100m/s and S. as a veriable on the y- axis. vo?(0.8660) 2)5. 675 = Hence Slope represents ( Jwill be constant. Vo =8.0137m/s The graph :| 10 3) 6. 17 = e@.9396) 8.0 Ve =8.0220 m/s Vo0.98480) 60 4) 6.435 = 4.0 Ve =8.0022m/s 5)6. 665 = 'o1) 2.0 Ve =8.0818 m/s 00 0.0 02 04 0.6 05 LO 12 14 16 LS 20 (8.0100+8.0137+8.0220+8.0022+8.0918) Average (initial velocity value )= (5) from the Range of angles from (25 - 45 ) there is a linear correlation as it is shown form the 40.1297 graph. = 8.0259 m/s , Hence Vo= 8.0259 m/s Slope is calculated automatically from the graph by this equation To check our work the slope must be equal to Vo = 8.0259 m/s - S,= (6.811).sin20 ( Slope is 6.811)

Sam has made this table for his experiment and he draw that graph below. - math calculation: information table (Data): Lunching angle 0 () (5.675-5.015) Range Trial 1 Range Trial 2 Average Range (m) * slope (trial 2) = = 6.6 (0.8660-0.7660) 20 Sin(20) s3 (m) s3 (m) 25 50 0.7660 5.03 5.00 5.015 Vosin20 To find Vg let's take any value for (Range value in the trial ) from this equation s, = 30 60 0.8660 5.66 5.69 5.675 35 70 0.9396 6.19 6.15 6.17 v.(0. 7660) 0.98480 6.435 40 80 6.44 6.43 1) 5.015 = 45 90 1 6.64 6.69 6.665 Vosin20 from this equation S, = , we can put that sin20 as a variable on the X- axis, Vo = 8.0100m/s and S. as a veriable on the y- axis. vo?(0.8660) 2)5. 675 = Hence Slope represents ( Jwill be constant. Vo =8.0137m/s The graph :| 10 3) 6. 17 = e@.9396) 8.0 Ve =8.0220 m/s Vo0.98480) 60 4) 6.435 = 4.0 Ve =8.0022m/s 5)6. 665 = 'o1) 2.0 Ve =8.0818 m/s 00 0.0 02 04 0.6 05 LO 12 14 16 LS 20 (8.0100+8.0137+8.0220+8.0022+8.0918) Average (initial velocity value )= (5) from the Range of angles from (25 - 45 ) there is a linear correlation as it is shown form the 40.1297 graph. = 8.0259 m/s , Hence Vo= 8.0259 m/s Slope is calculated automatically from the graph by this equation To check our work the slope must be equal to Vo = 8.0259 m/s - S,= (6.811).sin20 ( Slope is 6.811)

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based on sam's work write a short conclusion for his work with you general information about projectiles motion 

Sam has made this table for his experiment and he draw that graph below. - math calculation: information table (Data): Lunching angle 0 () (5.675-5.015) Range Trial 1 Range Trial 2 Average Range (m) * slope (trial 2) = = 6.6 (0.8660-0.7660) 20 Sin(20) s3 (m) s3 (m) 25 50 0.7660 5.03 5.00 5.015 Vosin20 To find Vg let's take any value for (Range value in the trial ) from this equation s, = 30 60 0.8660 5.66 5.69 5.675 35 70 0.9396 6.19 6.15 6.17 v.(0. 7660) 0.98480 6.435 40 80 6.44 6.43 1) 5.015 = 45 90 1 6.64 6.69 6.665 Vosin20 from this equation S, = , we can put that sin20 as a variable on the X- axis, Vo = 8.0100m/s and S. as a veriable on the y- axis. vo?(0.8660) 2)5. 675 = Hence Slope represents ( Jwill be constant. Vo =8.0137m/s The graph :| 10 3) 6. 17 = e@.9396) 8.0 Ve =8.0220 m/s Vo0.98480) 60 4) 6.435 = 4.0 Ve =8.0022m/s 5)6. 665 = 'o1) 2.0 Ve =8.0818 m/s 00 0.0 02 04 0.6 05 LO 12 14 16 LS 20 (8.0100+8.0137+8.0220+8.0022+8.0918) Average (initial velocity value )= (5) from the Range of angles from (25 - 45 ) there is a linear correlation as it is shown form the 40.1297 graph. = 8.0259 m/s , Hence Vo= 8.0259 m/s Slope is calculated automatically from the graph by this equation To check our work the slope must be equal to Vo = 8.0259 m/s - S,= (6.811).sin20 ( Slope is 6.811)
Transcribed Image Text:Sam has made this table for his experiment and he draw that graph below. - math calculation: information table (Data): Lunching angle 0 () (5.675-5.015) Range Trial 1 Range Trial 2 Average Range (m) * slope (trial 2) = = 6.6 (0.8660-0.7660) 20 Sin(20) s3 (m) s3 (m) 25 50 0.7660 5.03 5.00 5.015 Vosin20 To find Vg let's take any value for (Range value in the trial ) from this equation s, = 30 60 0.8660 5.66 5.69 5.675 35 70 0.9396 6.19 6.15 6.17 v.(0. 7660) 0.98480 6.435 40 80 6.44 6.43 1) 5.015 = 45 90 1 6.64 6.69 6.665 Vosin20 from this equation S, = , we can put that sin20 as a variable on the X- axis, Vo = 8.0100m/s and S. as a veriable on the y- axis. vo?(0.8660) 2)5. 675 = Hence Slope represents ( Jwill be constant. Vo =8.0137m/s The graph :| 10 3) 6. 17 = e@.9396) 8.0 Ve =8.0220 m/s Vo0.98480) 60 4) 6.435 = 4.0 Ve =8.0022m/s 5)6. 665 = 'o1) 2.0 Ve =8.0818 m/s 00 0.0 02 04 0.6 05 LO 12 14 16 LS 20 (8.0100+8.0137+8.0220+8.0022+8.0918) Average (initial velocity value )= (5) from the Range of angles from (25 - 45 ) there is a linear correlation as it is shown form the 40.1297 graph. = 8.0259 m/s , Hence Vo= 8.0259 m/s Slope is calculated automatically from the graph by this equation To check our work the slope must be equal to Vo = 8.0259 m/s - S,= (6.811).sin20 ( Slope is 6.811)
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