A student stands at the edge of a cliff and throws a stone horizontally over the
ID: 2135031 • Letter: A
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 = 49.0 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
Let: t be the time from the moment the stone is thrown, v1 be its horizontal velocity, v2 be its vertical velocity upwards, V be its speed on landing, x be the horizontal distance travelled, y be the vertical height above the landing point, g be the acceleration due to gravity, a be the angle of impact. Horizontally, the speed v0 is constant: x = v0 t Vertically, the initial velocity is 0. The acceleration is constant, downwards, and has magnitude g. y = h + 0t - gt^2 / 2 = h - gt^2 / 2 (a) Putting y = 0 gives the time of landing: 0 = h - gt^2 / 2 t^2 = 2h / g t = sqrt( 2 * 49.0 / 9.81 ) = 3.16 sec. (b) Horizontally: v1 = v0 = 16.0 m/s. Vertically: v2^2 = 0^2 - 2gh = - 2g(- h) = 2gh = 2 * 9.81 * 49.0 = 961.38 (m/s)^2 v2 = - sqrt(2 * 9.81 * 49.0) = - 31.0 m/s. (Negative root as v2 is downwards). The speed on landing is: V = sqrt(v1^2 + v2^2) sqrt(16.0^2 + 961.38) = 34.9 m/s. The angle of impact is given by: tan(a) = v2 / v1 = - 31.0 / 16.0 a = - 62.7 deg. = 62.7 deg. below the horizontal.
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