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# A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis). The initial velocity is 40 i + 64 k,?

with speed measured in feet per second. The spin of the ball results in a southward acceleration of 8 ft/s2, so the acceleration vector is a = −8 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.)

I have:

a(t)=<0,18,-32>

v(t)=<40,-8t,64-32t>

r(t)=<40t,-8t^2,64t-32t^s>

when the z component is zero t=0 and t=2

I plugged in t=2 into my position function and tried to do inverse tan to get the first two parts, but my answers are wrong. Can you help me see where I went wrong?

### 1 Answer

- Steve4PhysicsLv 75 months agoFavorite Answer
u = <40, 0, 64> (initial velocity)

a = <0, -8, -32> (you have a typing mistake)

v(t) = <40 + 0*t, 0 + (-8)t, 64 + (-32)t>

. . . .= <40, -8t, 64 – 32t> (agreed)

Remember displacement (s) in a given direction is: s= ut + ½at². You forgot the “½”:

r(t) = <40t, 0t + ½(-8)t², 64t + ½(-32)t²>

. . . = <40t, -4t², 64t – 16t²>

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The time in the air is the time for the vertical (z) velocity to change from 64 ft/s to -64ft/s.

v = u + at

-64 = 64 + (-32)t

t = 128/32 = 4s (I think you only worked out half the time, i.e. time to reach max. height)

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Since the ball starts at the origin (0,0,0) its final position is given by r(4):

r(4) = <40*4, -4(4)², 64*4 – 16(4)²>

. . . .= <160, -64, 0>

(note the x and y components tell us this is the 4th quadrant)

If θ is the angle south of east then:

θ = tan⁻¹(64/160) = 21.8º

Distance from origin = √(160² + (-64)²) = 172.3 ft

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v(4) = <40, -8*4, 64 – 32*4> = <40, -32, -64>

Speed = √(40² + (-32)² + (-64)²) = 82.0 ft/s

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