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An archerfish shoots at a target 2.00 m away, at an angle of 30.0o above the horizontal. With what velocity must the water stream be launched if it is not to drop more than 3.00 cm vertically on its path to the target?
i have troubles with this problem (Physics for scientists and engineers 6th edition)(Serway, Jewett)
i need to know what is wrong in my analysis
My Answer
target position x=2cos30=1.73 m. , y=2sin30=1 m.
maximum height H=1+0.03=1.03 m.
but H=vi2sin2θi/2g
=> vi2sin2θi=(9.8)(2.06)
the equation of the trajectory is y=(tanθi)x-[g/(2vi2cos2θi)]x2
=> 1=1.73tanθi-[(9.8)(1.5)/(vi2cos2θi)]
substituting for the value of vi2
we get a quadratic equation in tanθ
0.73tan2θi-1.73tanθi+1=0
either θi=54° or θi=45°
hence vi=5.6 m/s or vi=6.35 m/s
this solution is wrong , i don't know why?
the solution manual answer is 25.8 m/s with a different approach in the problem solving
so what is wrong in my solution
i have troubles with this problem (Physics for scientists and engineers 6th edition)(Serway, Jewett)
i need to know what is wrong in my analysis
My Answer
target position x=2cos30=1.73 m. , y=2sin30=1 m.
maximum height H=1+0.03=1.03 m.
but H=vi2sin2θi/2g
=> vi2sin2θi=(9.8)(2.06)
the equation of the trajectory is y=(tanθi)x-[g/(2vi2cos2θi)]x2
=> 1=1.73tanθi-[(9.8)(1.5)/(vi2cos2θi)]
substituting for the value of vi2
we get a quadratic equation in tanθ
0.73tan2θi-1.73tanθi+1=0
either θi=54° or θi=45°
hence vi=5.6 m/s or vi=6.35 m/s
this solution is wrong , i don't know why?
the solution manual answer is 25.8 m/s with a different approach in the problem solving
so what is wrong in my solution