30. *-/6 points SerPSET94.P.054. 0/6 Submissions Used My Notes Ask Your Teache A

Question

30. *-/6 points SerPSET94.P.054. 0/6 Submissions Used My Notes Ask Your Teache A farm truck moves due east with a constant velocity of 14.50 m/s on a limitless horizontal stretch of road. A boy riding on the back of the truck throws a can of soda upward (see figure) and catches the projectile at the same location on the truck bed, but 13.0 m farther down the road. (a) In the frame of reference of the truck, at what angle to the vertical does the boy throw the can? (b) What is the initial speed of the can relative to the truck? m/s (c) What is the shape of the can's trajectory as seen by the boy? a symmetric section of a parabola opening downward a straight line segment upward and then 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 a symmetric section of a parabola opening downward a straight line segment upward and then downward (e) In this observer's frame of reference, determine the initial velocity of the can. magnitude direction m/s o above the horizontal Need Help? ReadIt

Explanation / Answer

find the time of flight of the can by considering its horizontal motion

x = xo+ vo*t + (0.5)*a*t^2

13 m = (14.50 m/s)*t + 0

t = 0.896 s

(a)
For the boy to catch the can at the same location on the truck bed, he must throw it straight up, at 0° to the vertical.

(b)
For the free fall
  
yf = yi + v*yi*t + (1/2)a*y*t^2

0 = 0 + v yi (0.896 s) - (1/2)(9.8 m/s^2)(0.896 s)^2

vyi = 4.3904 m/s

(c)

The boy see the can always over his head, traversing a straight line segment upward and then downward.

(d)

The ground observer see the can move as a projectile on a symmetric section of a parabola opening downward.

(e)
Its initial velocity is

vo = sqrt(14.50^2+ 4.39^2)

= 15.149 m/s

tan^-1(4.39/14.50)

16.84 degree above the horizontal

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