Respuesta :
Answer:
The answer is "52.61 ft/s"
Explanation:
Taking the motion in the z-direction is [tex]u_z = 32 , \ \ a_z = - 32[/tex]
[tex]t = 2 \frac{u_z}{( - a_z )}\\\\[/tex]
[tex]= 2 \times \frac{32}{32}\\\\ = 2 \ sec\\\\[/tex]
z-axis change is zero z and the x-path the same way:
[tex]= u_x t + \frac{1}{2} a_x t^2 \\\\= 40 \times 2 + 0.5 \times 0 \times 2^2 \\\\ = 40 \times 2 + 0.5 \times 0 \times 4 \\\\ = 80 +0\\\\= 80 \ ft \ \ i[/tex]
In y-direction:
[tex]= 0 \times t - \frac{1}{2} \times (6) \times 2^2\\\\ = - \frac{1}{2} \times (6) \times 4\\\\ = - (6) \times 2\\\\= - 12 ft \ \ j[/tex]
the ball lands at:[tex]( 80 i - 12 j + 0 k ) ft[/tex]
Magnitude:
[tex]\to ( 80^2 + (-12)^2 )^{\frac{1}{2}}\\\\\to ( 6400 + 144 )^{\frac{1}{2}}\\\\\to ( 6544 )^{\frac{1}{2}}\\\\\to 80.89 \ ft[/tex]from the origin.
angle:
[tex]\to \tan^{-1} ( \frac{12}{80} )\\\\\to 8.53^{\circ}[/tex]
velocity:
[tex]\to 40 i + ( 0 - 6 \times 2 ) j + ( 32- 32 \times 2 ) k\\\\\to 40 i + ( - 12 ) j + ( 32- 64 ) k\\\\\to 40 i - 12 j - 32 k\\\\\to ( 40^2 + 12^2 + 32^2 )^{\frac{1}{2}}\\\\\to (1600+144+1024)^{\frac{1}{2}}\\\\\to 2768^{\frac{1}{2}}\\\\\to 52.61\ \ \frac{ft}{s}[/tex]
