Question

Exp 5 Projectile Motion Date: Name Part 1: Average Range & initial Velocity Height Time Velocity Number Ax (m) Average(m) Time Velocity (m/s) 1 1.98 2 2.01 $v_0 = \sqrt{\frac{L}{2H}} 3 2.06 2,04 4 2,07 5 2.08 Part 2 Distances per Assigned Angles Angle 30° 40° 45° 60° Horizontal Dist 1 2,90 2,91 2,89 2,28 Horizontal Dist 2 2,87 2,93 2,82 2.32 Horizontal Disys Average Distance Paper Distance Total Distance Calculated L $L = \frac{v_0^2 \sin 2\theta}{2g} \left(1 + \sqrt{1 + \frac{2gH}{v_0^2 \sin^2 \theta}}\right)$ Percent Error

          Exp 5 Projectile Motion
Date:
Name
Part 1: Average Range & initial Velocity
Height
Time Velocity
Number Ax (m)
Average(m) Time Velocity (m/s)
1
1.98
2
2.01
$v_0 = \sqrt{\frac{L}{2H}}
3
2.06
2,04
4
2,07
5
2.08
Part 2 Distances per Assigned Angles
Angle
30°
40°
45°
60°
Horizontal Dist 1
2,90
2,91
2,89
2,28
Horizontal Dist 2
2,87
2,93
2,82
2.32
Horizontal Disys
Average Distance
Paper Distance
Total Distance
Calculated L
$L = \frac{v_0^2 \sin 2\theta}{2g} \left(1 + \sqrt{1 + \frac{2gH}{v_0^2 \sin^2 \theta}}\right)$
Percent Error

Exp 5 Projectile Motion
Date:
Name
Part 1: Average Range initial Velocity
Height
Time Velocity
Number Ax (m)
Average(m) Time Velocity (m/s)
1
1.98
2
2.01
v0 = √((L)/(2H))
3
2.06
2,04
4
2,07
5
2.08
Part 2 Distances per Assigned Angles
Angle
30°
40°
45°
60°
Horizontal Dist 1
2,90
2,91
2,89
2,28
Horizontal Dist 2
2,87
2,93
2,82
2.32
Horizontal Disys
Average Distance
Paper Distance
Total Distance
Calculated LL = (v0^2 sin2θ)/(2g) (1 + √(1 + (2gH)/(v0^2 sin^2 θ)))Percent Error

Added by Paula C.

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