A player kicks up a ball at an angle 0 with horizontal. The horizontal range is maximum when '0' is equal to
Answers
Answer:
hilu ❤️
Explanation:
Angle of projection θ=30
Angle of projection θ=30 0
Angle of projection θ=30 0
Angle of projection θ=30 0 initial velocity, μ=15 m/sec
Angle of projection θ=30 0 initial velocity, μ=15 m/seca) Maximum height =
Angle of projection θ=30 0 initial velocity, μ=15 m/seca) Maximum height = 2g
Angle of projection θ=30 0 initial velocity, μ=15 m/seca) Maximum height = 2gμ
Angle of projection θ=30 0 initial velocity, μ=15 m/seca) Maximum height = 2gμ 2
Angle of projection θ=30 0 initial velocity, μ=15 m/seca) Maximum height = 2gμ 2 sin
Angle of projection θ=30 0 initial velocity, μ=15 m/seca) Maximum height = 2gμ 2 sin 2
Angle of projection θ=30 0 initial velocity, μ=15 m/seca) Maximum height = 2gμ 2 sin 2 θ
=
= 2×9.8
= 2×9.8(15)
= 2×9.8(15) 2
= 2×9.8(15) 2 ×sin
= 2×9.8(15) 2 ×sin 2
= 2×9.8(15) 2 ×sin 2 30
= 2×9.8(15) 2 ×sin 2 30 0
= 2×9.8(15) 2 ×sin 2 30 0
= 2×9.8(15) 2 ×sin 2 30 0
= 2×9.8(15) 2 ×sin 2 30 0
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 m
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile =
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ =
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.8
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 sec
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range =
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range = g
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range = gμ
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range = gμ 2
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range = gμ 2 sin2θ
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range = gμ 2 sin2θ
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range = gμ 2 sin2θ
= 2×9.8(15) 2 ×sin 2 30 0 =2.869 mb) Time of flight for projectile = g2μsinθ = 9.82×15×sin30 0 =1.53 secc) Range = gμ 2 sin2θ