what is the angle of projection to have maximum range in kitti pull? If one strikes kitti pull with the speed 98ms-1. what is the maximum range achieved
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R=(v2o∗sin(2θ))/gR=(vo2∗sin(2θ))/g
H=(v2o∗sin2(θ))/(2∗g)H=(vo2∗sin2(θ))/(2∗g)
vovo is the initial velocity
θθ is the angle of projection
gg is the gravity
RR is the range
HH is the height
Given that v_o is same and you want to know when R = 4H we do the following and solve for θ:
(v2o∗sin(2θ))/g=4∗((v2o∗sin2(θ))/(2∗g))(vo2∗sin(2θ))/g=4∗((vo2∗sin2(θ))/(2∗g))
which simplifies to:
sin(2θ)=2∗sin2(θ)sin(2θ)=2∗sin2(θ)
I hope.
Edit 2018–09–26:
Brian Alan Whatcott suggested I simplify the sin(2θ)sin(2θ) bit in terms of θθ, instead of that here are a list of trigonometric identities which state that simplification.
H=(v2o∗sin2(θ))/(2∗g)H=(vo2∗sin2(θ))/(2∗g)
vovo is the initial velocity
θθ is the angle of projection
gg is the gravity
RR is the range
HH is the height
Given that v_o is same and you want to know when R = 4H we do the following and solve for θ:
(v2o∗sin(2θ))/g=4∗((v2o∗sin2(θ))/(2∗g))(vo2∗sin(2θ))/g=4∗((vo2∗sin2(θ))/(2∗g))
which simplifies to:
sin(2θ)=2∗sin2(θ)sin(2θ)=2∗sin2(θ)
I hope.
Edit 2018–09–26:
Brian Alan Whatcott suggested I simplify the sin(2θ)sin(2θ) bit in terms of θθ, instead of that here are a list of trigonometric identities which state that simplification.
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