A ball is thrown eastward into the air from the origin (in the direction of the

Question

A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is

50 i + 32 k,

with speed measured in feet per second. The spin of the ball results in a southward acceleration of 6 ft/s2, so the acceleration vector is a = 6 j 32 k. Where does the ball land? (Round your answers to one decimal place.)

____ ft from the origin at an angle of____ ° from the eastern direction toward the south.

With what speed does the ball hit the ground? (Round your answer to one decimal place.)

__

Will rate

Explanation / Answer

Given that

The initial velocity is (vo) = 50 i + 32 k,

Thn the acceleration vector is (a) = 6 j 32 k.

to find the final velocity of ball after a time t seconds we have to integrate the term acceleration

               integeration of a we get v(t) = -6tj-32tk+C

Now weknow that 50 i + 32 k, =-6(0)j-32(0)k+C at t =0s

The cosntantC =50i+32k

Now velocity vector becomes (v) = 50i -6tj+(32-32t)k.

integrating to find the position vector then we get

r(t) =integration of pv(t)dt

     =integration of 50i -6tj+(32-32t)k.

     =50ti-3t2j+(32t-16t2)k +c

Here ball reaches the ground in southward then k- component is zero

    0=50ti-3t2j at t =0seconds t =(1/2)

The position vector r(1/2) =25i-(3/4)j =25i-(0.7)j

v(1/2) =50i -6(1/2)jtj+32(1-1/2)k

         = 50i-3tj+16k

The speed does the ball hit the ground is v = Sqrt[(50)2+(-3)2+(16)2] =52.58m/s

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