A farm truck moves due east with a constant velocity of 14.00 m/s on a limitless

Question

A farm truck moves due east with a constant velocity of 14.00 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 below) and catches the projectile at the same location on the truck bed, but 14.0 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? (e) In this observer's frame of reference, determine the initial velocity of the can. magnitude______ m/s direction_______

Explanation / Answer

(a) Relative to the truck, the initial speed of the can can be determined by, first, determining how long the can was in the air for: 14m/x=10m/s x=1.4s So the can was in the air for 1.4s. Knowing this, and assuming the acceleration of gravity is 9.81m/s^2, you can find the relative initial speed using: D=Vi(t)+1/2a(t)^2 You know that the net distance (again, relative to the truck) was 0. Now you know everything you need to solve for initial velocity: 0=Vi(1.4)+1/2(-9.81)(1.4)^2 4.905(1.4)=Vi Vi=6.87m/s So the relative initial speed was approximately 6.87 m/s. (b) With respect to the observer, the vector in which the can travels on is different. To determine the magnitude of the can, we must use the pythagorean theorem: x^2+y^2=z^2 10^2+6.87^2=z^2 v(100+47.16)=z z=v147.16 z=12.13 m/s As for the direction, all you must do is divide the rise by the run and take the arctangent of that fraction: 6.87/10=tan(?) .687=tan(?) ?=arctan(.687) ?=34.5° Hope this helps. Source(s): High school physics

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