the record distance in the sport of throwing cowpats is 81.1 m. this record toss was set by steve urner of the united states in 1981. assuming the initial launch angle was 45° and neglecting air resistance, answer the following.

a) When a projectile is launched with speed [tex]v_{i}[/tex] at an angle above the horizontal, the initial velocity components are [tex]v_{xi}[/tex] = [tex]v_{i}[/tex] cos[tex]θ[/tex] and [tex]v_{yi}[/tex]= [tex]v_{i}[/tex] sin[tex]θ[/tex] . Neglecting air resistance, the vertical velocity when the projectile returns to the level from which it was launched (in this case, the ground) will be [tex]v_{y} = - v_{yi}[/tex] .

From this information, the total time of flight is found [tex]v_{yf} = v_{yi} + a_{y}[/tex] to be

[tex]t_{total} = \frac{v_{yf} - v_{yi} }{a_{y}}[/tex] = [tex]\frac{-v_{yi} - v_{yi} }{-g}}[/tex] = [tex]\frac{2v_{yi} }{g}[/tex]

[tex]t_{total} = \frac{2v_{i}sinθ }{g}[/tex]

Since the velocity of a projectile with no air resistance is constant, the horizontal distance it will travel in this time is given by

R = [tex]v_{xi} t_{total} = (v_{i} cosθ_{i}) \frac{2v_{i}sinθ }{g}[/tex] = [tex]\frac{v^{2}_{i}}{g}( 2sinθ_{i} cosθ_{i} )[/tex] = [tex]\frac{v^{2} _{i} (sin2θ_{i} ) }{g}[/tex]

Thus, if the projectile is to have a range of R=81.1m when launched at an angle of [tex]θ_{i}[/tex]=45.0°, the required initial speed is

[tex]v_{i} = \sqrt{\frac{81.1 * 9.80}{sin90} }[/tex] = 28.2 m/s

Therefore, the initial speed of the projectile - 28.2m/s

b) With [tex]v_{i}[/tex]=28.2m/s and [tex]θ_{i}[/tex] = 45.0°, the total time of flight (as found

above) will be

[tex]t_{total} = \frac{2v_{i}sinθ }{g}[/tex] = [tex]\frac{2(28.2 m/s) sin45}{9.80m/s^{2}}[/tex] = 4.07s

∴The total time interval the projectile was in flight - was 4.07s.

c) Note that at [tex]θ_{i}[/tex] =45.0° that sin(2[tex]θ_{i}[/tex]) will decrease as [tex]θ_{i}[/tex] is increased above this optimum launch angle. Thus, if the range is to be kept constant while the launch angle is increased above 45.0°, we see from [tex]v_{i}[/tex] = [tex]\sqrt{\frac{Rg}{sin2θ_{i}} }[/tex]that the required initial velocity will increase.

Observe that for [tex]θ_{i}[/tex] <90°, the function sin[tex]θ_{i}[/tex] increases as [tex]θ_{i}[/tex] it increased. Thus, increasing the launch angle above 45.0° while keeping the range constant means that both [tex]v_{i}[/tex] sins will increase. Considering the expression [tex]t_{total}[/tex] given above, we see that the total time of flight will increase.