A physics student is standing on the back of a flatbed truck tossing balls verti

Question

A physics student is standing on the back of a flatbed truck tossing balls vertically up into the air with velocity relative to the truck of Vbt. The truck starts from rest and moves with a constant acceleration of at. The instant the truck starts, the physics tosses mother ball (NO wind resistance).

Assuming the ball lands on the bed of the truck, how far from the student feet does it land? That is what is the final position of the ball relative to the truck or student?

Please show work and explain cleary for I am very lost. This is for an exam so I need to understand. Also please draw a clear diagram.  

Explanation / Answer

let S be the distance where the ball lands

we know that

Vbt^2 - u^2 = 2at x S

or S = (Vbt^2 - u^2/2at)

where u = 0(the ball starts from rest) and at is the acceleration

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