A student stands at the edge of a cliff and throws a stone horizontally over the
ID: 1326890 • Letter: A
Question
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of
The cliff is h = 35.0 m above a flat, horizontal beach as shown in the figure.
(a) What are the coordinates of the initial position of the stone?
(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 include 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?
?sec
(f) With what speed and angle of impact does the stone land ?
x0 =? mExplanation / Answer
Given that,
Vo = 16.5 m/s and h = 35 meters
(a)The coordinates of initial position of the stone is:
(x0 , y0) = (0,0).
(b) components of the initial velocity is
(Vox, Voy) = (16.5 m/s, 0 ).
(c)The x component of velocity will be same as intial and is independent of time. So
Vx = 16.5
Vy = Voy + ay t = 0 - 9.8 t = -9.8 t
Vy = -9.8 t
(d)
X = Vox t + 1/2 ax t2 , but ax = 0
X = 16.5 t
Y = Voy t + 1/2 ay t2 , but Voy = 0 and ay = -9.8, So
Y = -1/2 x 9.8 x t2 = - 4.9 t2
Y = - 4.9 t2
(e)We have, Y = - 4.9 t2 ( Y = h = 35 m)
-4.9 t2 = - 35 => t = 2.67 sec
Hence, t = 2.67 sec
(f)We have, Vx = 16.5 m/s
and as calculated in the part c and e , we have, Vy = -9.8 t and t = 2.67 sec
Vy = -9.8 x 2.67 = -26.17 m/s
Vf = sqrt [(Vx)2 + (Vy)2] = sqrt [(16.5)2 + (-26.17)2] = 31 m/s
= tan -1 ( -26.17/16.5) = 55.82° below the horizontal
Hence, Vf = 34.71 and = 57.77° below the horizontal.
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