(a) The lunch angle is 64.8⁰
(b) The initial speed of the pass when the angle of projection is 25⁰ is 21.2 m/s
(c) The time of flight of the bullet is 1.83 s
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The given parameters include;
time of flight, T = 3.97 s
initial velocity, u = 21.5 m/s
(a) The lunch angle is calculated from the equation of motion of time of flight;
[tex]T = \frac{2u sin(\theta)}{g} \\\\sin(\theta ) = \frac{Tg}{2u} \\\\sin(\theta ) = \frac{3.97 \times 9.8}{2 \times 21.5} \\\\sin(\theta ) = 0.905\\\\\theta = sin^{-1} (0.905)\\\\\theta = 64.8 ^0[/tex]
(b) the initial speed of the pass when the angle of projection is 25⁰ and range of 35 m, is calculated as follows;
[tex]R = \frac{u^2 sin (2\theta )}{g} \\\\u^2 = \frac{Rg}{sin(2\theta )} \\\\u = \sqrt{\frac{Rg}{sin(2\theta )}}\\\\ u = \sqrt{\frac{35 \times 9.8}{sin(2 \times 25 )} }\\\\u = 21.2 \ m/s[/tex]
(c) The time of flight of the bullet is calculated as;
[tex]T = \frac{2u sin(\theta )}{g} \\\\T = \frac{2\times 21.2 \times sin(25)}{9.8} \\\\T = 1.83 \ s[/tex]
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