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-14 points SAIPSE9 4 P010 My Noles Ask Your Tes A snowmobile is originally at the point with position vector 30.1 m at 95.09 counterclockwise from the x axis, moving with velocity 4.89 m/s at 40.0°. It moves with constant acceleration 1.94 m/s at 200° After 5.00 s have elapsed, fina the following. (a) its velocity vector (b) its position vector Submit Aeswer Save Progress Pracice Anoher Versice A ball is tossed from an upper-story window af a building. The ball is given an initial valocity ot 9.10 m/s at an angle of 17.0° belows the horizontal. It strikes the ground 6.00 s later (a) How far horizontally from the bise of the bualding does the ball stnke the ground?

Explanation / Answer

the velocity vector in x direction is vx'= vx+ax* time = vcost+acost *time = 4.89cos40+1.94os200*5=-5.36i

in the vertical direction vy'= vy+ay* time = vsint+asint *time = 4.89sin40+1.94*sin200*5 =-0.17j

so the velocity ecto v=-5.36i-0.17j

where p is the time
the position vector in the horiontal direction x'=x0+vcost*p+0.5*acost*p^2 = xcos95+vcost*p+0.5*acost*p^2= -6.67i

the position vector in the vertical direction y'=y0+vsint*p+0.5*asint*p^2 = xsin95+vsint*p+0.5*asint*p^2= 37.40j

so the position vector is -6.67i+37.40j

2) the horizontal distance ball falls is x =vcos(t)* time = 9.10*cos(17)*6 = 52.214 m

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