A stone is thrown upward from the edge of a cliff, reaches its maximum height, a

Question

A stone is thrown upward from the edge of a cliff, reaches its maximum height, and then falls down into the valley below. A motion diagram for this situation is given , beginning the instant the stone leaves the thrower?s hand. Construct the corresponding motion graphs taking the acceleration due to gravity as 10 ay(t). vy(t) c-Construct a graph corresponding to the stone's vertical acceleration, y(t). b-Construct a graph corresponding to the stone's vertical velocity, m/s^2. Ignore air resistance. In all three motion graphs, the unit of time is in seconds and the unit of displacement is in meters. In plotting the points, round-off the coordinate values to the nearest integer. a-Construct a graph corresponding to the stone's vertical displacement,

Explanation / Answer

1) The vertical displacement equation is: yf(t) = (vi * t) + ((-10 * t^2)/2) Remember that x=0 and x=4 both have y=0... if that makes any sense. Also, vi=20 in this case. Make sure you account for 't' when calculating y(f). 2) The vertical velocity is the vertical displacement divided by the period of time ((yf - yi)/(tf - ti)) Hint: yf=-60 ; yi=0; tf=6; ti=0 3) The acceleration is merely the change in velocity over a period of time.

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