Ok so I have been trying this problem and I have been unable tofind the final da

Question

Ok so I have been trying this problem and I have been unable tofind the final dart distance. My question states"A dart gun isfired while being held horizontally at a heightof 1.10 m above ground level andwhile it is at rest relative to the ground. The dart from the guntravels a horizontal distance of 4.90 m. A college student holds the same gun in ahorizontal position while sliding down a 45.0° incline at aconstant speed of 2.00 m/s. Howfar will the dart travel if the student fires the gun when it is1.00 m above the ground?"
I keptfinding the time to be 0.4518 sec and the final distance to be4.673m. Can someone please hep me find the correctanswer?
I keptfinding the time to be 0.4518 sec and the final distance to be4.673m. Can someone please hep me find the correctanswer?

Explanation / Answer

Given : Height ( h ) = 1.10 m   distance ( d ) = 4.90 m/s g = 9.8 m/s2 Time ( t ) = 2h/g                  = 2*1.10 / 9.8                 = 0.4738 s Distance ( d ) = v * t    v = d / t      = 4.90 m/s / 0.4738 s      = 10.34 m/s for 2nd person t = 2h / g in     = 2*1.10 / 9.8 sin45     = 0.57 s Now the horizontal distance    distance ( x ) = v * t                        = ( 2m/s ) ( 0.57 s ) = 1.13 m Final distance = v * t = ( 10.34 m/s ) * ( 0.57 s ) =5.9 m I hope it helps you Time ( t ) = 2h/g                  = 2*1.10 / 9.8                 = 0.4738 s Distance ( d ) = v * t    v = d / t      = 4.90 m/s / 0.4738 s      = 10.34 m/s for 2nd person t = 2h / g in     = 2*1.10 / 9.8 sin45     = 0.57 s Now the horizontal distance    distance ( x ) = v * t                        = ( 2m/s ) ( 0.57 s ) = 1.13 m Final distance = v * t = ( 10.34 m/s ) * ( 0.57 s ) =5.9 m I hope it helps you

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