A student stands at the edge of a cliff and throws a stone horizontally over the
ID: 1278729 • Letter: A
Question
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 12.0 m/s. The cliff is h = 42.0 m above a flat, horizontal beach as shown in the figure.
(a) What are the coordinates of the initial position of the stone? 0, 42
(b) What are the components of the initial velocity?
(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.) HAVE TRIED BUT CAN"T GET THE EQUASIONS CORRECT!
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?
2.93 s
(f) With what speed and angle of impact does the stone land?
Vf= 2.87 (incorrect)
O= 72.01 (incorrect)
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s
Explanation / Answer
a) x = 0 y = 42
b) v0x = 12
v0y = 0
c) vx = v0x = 12
vy = v0y - g t = -9.81 t
d) x = v0x t = 12 t
y = y0 - 1/2 g t^2 = 42 - 4.9 t^2
e)
so y = 0
0 = 42 - 4.9 t^2
t=2.93 s
f) vx = 12
vy = -9.81*2.93=-28.74
v = sqrt(12^2 + 28.74^2)= 31.14 m/s
angle = arctan(vy/vx) = arctan(38.74/12)= 72.8 degrees
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