y= x (sinb / cosb ) - (gx² / 2u² cos² b)
In the above equation, put y = 0. u and x is given. Rewrite the above equation.
0 = x*tanθ -(gx^2/2u^2)sec^2(θ)
Substitute sec^2(θ) = 1 + tan^2(θ) and solve the quadratic to find the angle.
I have this question and I have no idea where to start it. Any help would be extremely appreciated!
If a ball is hit into the air at an angle of elevation between 0 and 90 degrees its path can be modelled by the equation:
y= x (sinb / cosb ) - (gx² / 2u² cos² b)
where x and y are cartestian coordinates.
g=10m/s² (gravity), u = (initial velocity), b = the angle of projection. Lyall can hit a cricket ball at 40m/s. The boundary at the MCG is 70m from the pitch.
Is he capable of hitting a six and if so calculate the angle at which the ball should be hit to clear the boundary rope? (i.e. y=0 and x=0)
Honestly, I don't know where to start and don't know which rules to apply.
Please help