A farm truck moves due north with a constant velocity of10.00 m/s on a limitless

Question

A farm truck moves due north with a constant velocity of10.00 m/s on a limitless horizontalstretch of road. A boy riding on the back of the truck throws a canof soda upward and catches the projectile at the same location onthe truck bed, but 15.0 m farther downthe road.

(a) In the frame of reference of the truck, at what angle to thevertical 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 theboy?
3 - a symmetricsection of a parabola opening downward
    - a straightline segment upward and then downward

(d) An observer on the ground watches the boy throw the can andcatch it. In this observer's ground frame of reference, describethe shape of the can's path and determine the initial velocity ofthe can.
4 - a straight linesegment upward and then downward
    -a symmetric section of aparabola opening downward

5    _____________m/s north at 6__________°above the horizontal
(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 theboy?
3 - a symmetricsection of a parabola opening downward
    - a straightline segment upward and then downward

(d) An observer on the ground watches the boy throw the can andcatch it. In this observer's ground frame of reference, describethe shape of the can's path and determine the initial velocity ofthe can.
4 - a straight linesegment upward and then downward
    -a symmetric section of aparabola opening downward

5    _____________m/s north at 6__________°above the horizontal 3 - a symmetricsection of a parabola opening downward
    - a straightline segment upward and then downward
3 - a symmetricsection of a parabola opening downward
    - a straightline segment upward and then downward
3 4 - a straight linesegment upward and then downward
    -a symmetric section of aparabola opening downward
4 - a straight linesegment upward and then downward
    -a symmetric section of aparabola opening downward
4

Explanation / Answer

a) 0 degrees fromvertical b)To find the velocity, first solve for time. No horizontal acceleration, so Dx =vxt 15 = 10t t = 1.5 seconds Now solve for the vertical velocity y = vot + (1/2)at2 0 = vo(1.5) -(9.8/2)(1.5)2 vo = 7.35m/s c)a straight line segment upward and thendownward d)asymmetric section of a parabola openingdownward We know the horizontal and vertical components of velocityalready v = (vy2 +vx2) vy = 7.35 m/s vx = 10 m/s v = 12.4m/s To find the angle tan = vy/vx = tan-1(vy/vx) =tan-1(7.35/10) = 36.3 degrees abovehorizontal = tan-1(vy/vx) =tan-1(7.35/10) = 36.3 degrees abovehorizontal

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