A student stands at the edge of a cliff and throws a stone horizontally over the

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 14.0 m/s. The cliff is h = 58.0 m above a flat, horizontal beach as shown in the figure.

(a) What are the coordinates of the initial position of the stone?
x0 = m
y0 = m


(b) What are the components of the initial velocity?
v0x = m/s
v0y = m/s

(c) Write the equations for the x- and y-components of the velocity of the stone with time. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.)
vx =
vy =

(d) Write the equations for the position of the stone with time, using the coordinates in the figure. (Use the following as necessary: t. Let the variable t be measured in seconds. Do not state units in your answer.)
x =
y =


(e) How long after being released does the stone strike the beach below the cliff (s)?

(f) With what speed and angle of impact does the stone land?
vf = m/s
? =

Explanation / Answer

As is customary, we ignore air resistance in this kind of problem. The horizontal velocity of the stone is irrelevant for purposes of determining when the stone hits the ground. It's the same as a stone just dropped off the cliff. Use the distance formula d = 1/2 * a * t^2 + v0*t + d0 to find out how long it takes to drop. d is equal to 68m of course, and a is the acceleration of gravity, about 10m/sec^2. You will be able to find the vertical component of velocity by the formula v = a*t + v0 The horizontal velocity is assumed to be constant at 14 m/S, and when you know the vertical component of velocity, you can apply trigonometery, and take an arc tangent to get an angle. To get the speed, add the horizontal and vertical vectors. If you draw it out, you'll see that you can use the Pythagorean theorem to get the speed, which is a hypotenuse.

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