Answer:
Formulas:
T = v sin ∅ ÷ g
Tt = 2t
dy = v² sin² ∅ ÷ 2g
dx = v² sin (2∅) ÷ g
Explaination:
T = (10m/s) sin (30°) ÷ 9.8m/s²
T = 0.51s
Tt = 2 (0.51s)
Tt = 1.02s
dy = (10m/s)² sin² (30°) ÷ 2 (9.8m/s²)
dy = 25m²/s² ÷ 19.6m/s²
dy = 1.27m
dx = (10m/s)² sin (2 • 30°) ÷ 9.8m/s²
dx = 86.6m²/s² ÷ 9.8m/s²
dx = 8.83m
T = (10m/s) sin (45°) ÷ 9.8m/s²
T = 0.72s
Tt = 2 (0.72s)
Tt = 1.44s
dy = (10m/s)² sin² (45°) ÷ 2 (9.8m/s²)
dy = 50m²/s² ÷ 19.6m/s²
dy = 2.55m
dx = (10m/s)² sin ( 2 • 45°) ÷ (9.8m/s²)
dx = 100m²/s² ÷ 9.8m/s²
dx = 10.2m
T = (10m/s) sin (50°) ÷ (9.8m/s²)
T = 7.66m/s ÷ 9.8m/s²
T = 0.78s
Tt = 2 (0.78s)
Tt = 1.56s
dy = (10m/s)² sin² (50°) ÷ 2 (9.8m/s²)
dy = 58.68m²/s² ÷ 19.6m/s²
dy = 2.99m
dx = (10m/s)² sin ( 2 • 50°) ÷ (9.8m/s²)
dx = 98.48m²/s² ÷ 9.8m/s²
dx = 10.04m
T = (10m/s) sin (60°) ÷ (9.8m/s²)
T = 8.66m/s ÷ 9.8m/s²
T = 0.88s
Tt = 2 (0.88s)
Tt = 1.76s
dy = (10m/s)² sin² (60°) ÷ 2 (9.8m/s²)
dy = 74.99m²/s² ÷ 19.6m/s²
dy = 3.82m
dx = (10m/s)² sin (2 • 60°) ÷ (9.8m/s²)
dx = 86.6m²/s² ÷ 9.8m/s²
dx = 8.83m
Symbolism:
V = initial velocity
T = time ascending
Tt = Total time
dy = height
dx = range
