- #1
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A slug is shot at a 50 degree angle from a launcher 1.15m off the ground and has a muzzle velocity of 3.37m/s. Find the range of the slug.
I solved it this way but I'm not sure if I did it correctly.
Vvi=(sin50)(3.37m/s)=2.581m/s
Vhi=(cos50)(3.37m/s)=2.166m/s
[tex] vertical:[/tex]
[tex]v_i=2.581\frac{m}{s}[/tex]
[tex]d=1.15m[/tex]
[tex]a=g=9.81\frac{m}{s^2}[/tex]
[tex]t=?[/tex]
[tex]d=v_it+\frac{1}{2}at^2[/tex]
[tex]1.15=2.581t+4.905t^2[/tex]
[tex]4.905t^2+2.581t-1.15=0[/tex]
[tex]t=\frac{-2.581+\sqrt{6.661561+22.563}}{9.81}[/tex]
[tex]t=\frac{2.82497}{9.81}=.288s[/tex]
[tex]horizontal:[/tex]
[tex]t=.288s[/tex]
[tex]v_i=2.166\frac{m}{s}[/tex]
[tex]a=0[/tex]
[tex]d=?[/tex]
[tex]d=v_it[/tex]
[tex]d=(2.166)(.288)=.6238m[/tex]
so my answer is .6238m...but this doesn't seem correct at all. During the lab the slug went an average of 1.4 m each trial. Someone help me!
I solved it this way but I'm not sure if I did it correctly.
Vvi=(sin50)(3.37m/s)=2.581m/s
Vhi=(cos50)(3.37m/s)=2.166m/s
[tex] vertical:[/tex]
[tex]v_i=2.581\frac{m}{s}[/tex]
[tex]d=1.15m[/tex]
[tex]a=g=9.81\frac{m}{s^2}[/tex]
[tex]t=?[/tex]
[tex]d=v_it+\frac{1}{2}at^2[/tex]
[tex]1.15=2.581t+4.905t^2[/tex]
[tex]4.905t^2+2.581t-1.15=0[/tex]
[tex]t=\frac{-2.581+\sqrt{6.661561+22.563}}{9.81}[/tex]
[tex]t=\frac{2.82497}{9.81}=.288s[/tex]
[tex]horizontal:[/tex]
[tex]t=.288s[/tex]
[tex]v_i=2.166\frac{m}{s}[/tex]
[tex]a=0[/tex]
[tex]d=?[/tex]
[tex]d=v_it[/tex]
[tex]d=(2.166)(.288)=.6238m[/tex]
so my answer is .6238m...but this doesn't seem correct at all. During the lab the slug went an average of 1.4 m each trial. Someone help me!