How can the rock land "directly below the throwing point" and a horizontal distance x away?
Perhaps you left out some important words of the question.
A rock thrown horizontally from a bridge hits the water below at a horizontal distance x directly below the throwing point. The rock travels a smooth parabolic path in time t.
A.) Show that Therapy vertical distance of the bridge above water is 1/2gr2.
B.) What is the height of the bridge if the time of the rock is airborne is 2 seconds?
1 answer
1 answer