+33661 For the horizontal component of motion the velocity is constant, so if the initial angle is θ, we have, using distance = velocity* time, and that 1ft/s = 0.3048m/s, that:
2000 = 670.56*cos(θ)*t ...(1) where t is the flight time.
For the vertical motion we have vertical distance = initial velocity*time + (1/2)*accn*time2
Now the net vertical distance here is zero, so:
0 = 670.56*sin(θ)*t - (1/2)*9.8*t2 ...(2)
Rearrange equation (1) to get t = 2000/(670.56*cos(θ)) and substitute this into (2):
0 = 670.56*sin(θ)*2000/(670.56*cos(θ)) - (1/2)*9.8*(2000/670.56*cos(θ))2
Multiply this through by cos(θ)2 and divide through by 2000 to get:
0 = cos(θ)*sin(θ) - (1/2)*9.8*2000/670.562
or 0 = (1/2)sin(2θ) - (1/2)*9.8*2000/670.562
Multiply through by 2 and add 9.8*2000/670.562 to both sides:
sin(2θ) = 9.8*2000/670.562
so 2θ = sin-1(9.8*2000/670.562)
or θ = sin-1(9.8*2000/670.562)/2
$${\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{9.8}}{\mathtt{\,\times\,}}{\mathtt{2\,000}}}{{{\mathtt{670.56}}}^{{\mathtt{2}}}}}\right)}}{{\mathtt{2}}}} = {\mathtt{1.249\: \!139\: \!800\: \!311\: \!5}}$$
θ ≈ 1.25°. A very small angle!
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