a farm truck moves due east with a constant velocity of 14.5 m/s on a limitless,

Question

a farm truck moves due east with a constant velocity of 14.5 m/s on a limitless, horizontal stretch of road. a boy riding on the back of the truck throws a can of soda upward and catches the projectile at the same location on the truck bed, but 11 meters farther down the road. in the frame of reference of the truck, at what angle to the vertical does the boy throw? what is the initial speed of the can relative to the truck? In this observers frame of reference, determine the initial velocity of the can. Magnitude and direction a farm truck moves due east with a constant velocity of 14.5 m/s on a limitless, horizontal stretch of road. a boy riding on the back of the truck throws a can of soda upward and catches the projectile at the same location on the truck bed, but 11 meters farther down the road. in the frame of reference of the truck, at what angle to the vertical does the boy throw? what is the initial speed of the can relative to the truck? In this observers frame of reference, determine the initial velocity of the can. Magnitude and direction a farm truck moves due east with a constant velocity of 14.5 m/s on a limitless, horizontal stretch of road. a boy riding on the back of the truck throws a can of soda upward and catches the projectile at the same location on the truck bed, but 11 meters farther down the road. in the frame of reference of the truck, at what angle to the vertical does the boy throw? what is the initial speed of the can relative to the truck? In this observers frame of reference, determine the initial velocity of the can. Magnitude and direction

Explanation / Answer

(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 of the can,

yf = yi + vyit + (1/2)ayt2:

0 = 0 + vyi (0.759) - (1/2)(9.8 m/s2)(0.759 s)2

vyi= 3.71 m/s

The ground observer sees the can move as a projectile on

a symmetric section of a parabola opening downward.

Its initial velocity is:

v o=v(14.5^2+ 3.71^2) = 15 m/s north

at tan-1(3.71/14.5) = 14.4° above the horizontal

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