cowww.weba Assignment Responses/last?dep- 18065798 A farm truck moves due east w

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cowww.weba Assignment Responses/last?dep- 18065798 A farm truck moves due east with a constant velocity of 14.50 m/s on a limitles catches the projectile at the same location on the truck bed, but 13.0 m farther down the road. boy riding on the back of the truck throws a can of soda upward (see figure) ) In the frame of reference of the truck, at what angle to the vertical does the boy throw the can What is the initial speed of the can relative to the track? m/s (c) What is the shape of the can's trajectory as seen by the boy? a straight line segment upward and then downward a symmetric section of a parabola opening downward An observer on the ground watches the boy throw the can and catch it (d) In this observer's frame of reference, describe the shape of the can's path O a symmetric section of a parabola opening doenward a straight line segment upward and then downward (e) In this observer's frame of reference, determine the initial velocity of the can m/s o above the herizontal direction Need Help? v

Explanation / Answer

Observer's reference frame: The soda is in the air during the time it takes the truck to move forward 13.0 m

x = v t t = x/v
t = (13.0 m)/(14.50 m/s) = 0.896 s

Truck's frame: The soda was in flight for 0.896 s. It will land exactly where it was thrown from so y = 0m

y = v t - (1/2)g t²
(0 m) = v (0.896 s) - (1/2)(9.81m/s²)(0.896 s)²

v = (0.5)(9.81 m/s²)(0.896)
v = 4.395 m/s

We'll just call it v = 4.395 m/s to be consistent with our sig figs.

x-direction:
v = 14.50 m/s

y-direction
v = 4.395 m/s
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Part a:
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Our velocity vector will make a triangle whose angle we can calculate using the fact that the ratio of these two is the tangent of the angle between them.

(4.395 / 14.50) = tan()
= arctan(4.395 / 14.50)
16.86°
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Part b:
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We already used that to get part a. It is moving 0 m/s (relative to the truck) in the x-direction, and 4.395 m/s in the y.
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Part e:
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|V|² = (4.395 m/s)² + (14.50 m/s)²
|V| = 15.15 m/s

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