A farm truck moves due north with a constant velocity of9.00 m/s on a limitless

Question

A farm truck moves due north with a constant velocity of9.00 m/s on a limitless horizontal stretchof road. A boy riding on the back of the truck throws a can of sodaupward and catches the projectile at the same location on the truckbed, but 19.0 m farther down theroad. (a) In the frame of reference of the truck, at what angle tothe vertical does the boy throw the can?
(b) What is the initial speed of the can relative to the truck?
(c) What is the shape of the can's trajectory as seen by theboy?
(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.
A farm truck moves due north with a constant velocity of9.00 m/s on a limitless horizontal stretchof road. A boy riding on the back of the truck throws a can of sodaupward and catches the projectile at the same location on the truckbed, but 19.0 m farther down theroad. (a) In the frame of reference of the truck, at what angle tothe vertical does the boy throw the can?
(b) What is the initial speed of the can relative to the truck?
(c) What is the shape of the can's trajectory as seen by theboy?
(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.

Explanation / Answer

a) As far as the boy on the truck is concerned theobject was thrown straight up, unless wind resistance was takeninto account but it isn't in this problem. b) Well the initial horizontal speed is zero. the vertical initial speed is a little more complicated. First we find the time that the can was in the air d=vt t=d/v =(19 m/s)/(9 m) =2.11 sec Now we plug into the equation: df=di + vit +(1/2)at2 Where di & df are the initial andfinal height and a is the accelleration of gravity Where di & df are the initial andfinal height and a is the accelleration of gravity (0 m)=(0 m) + 2.11vi + (1/2)(-9.8m/s2)(2.112 sec) vi=10.34 m/sstraight up c)A straight line d)Well the initial verticle speed is still the same, but nowthe horizontal speed will be 9 m/s We can find the speed with the pythagoreum therom since theverticle and horizontal make a right angle v=(10.342+92) v=13.71m/s The angle can be found using the arc tangent tan=vy/vx =tan-1(vy/vx) =tan-1(10.34/9) =48.960above the horizontal =tan-1(vy/vx) =tan-1(10.34/9) =48.960above the horizontal

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