8:29 PM webassign.net AT&T; LTE A student stands at the edge of a cliff and thro
ID: 1654872 • Letter: 8
Question
8:29 PM webassign.net AT&T; LTE A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 25.0 ms. The cliff is n*S8.5 m above a body of water as shown in the figure below. (a) What are the coordinates of the initial position of the stone? Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m (b) What are the components of the initial velocity of the stone? (c) What is the appropriate analysis model for the vertical motion of the stone? (d) What is the appropriate analysis model for the horizontal motion of the stone? (e) Write symbolic equations for the x and y components of the velocity of the stone as a function of time. (Use the following as necessary:Vx g, and t. Indicate the direction of the velocity with the sign of your answer.) 9.81 x (f) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: Vixe 9, and t. Indicate the direction of the displacement with the sign of your answer.) 25 4.9Explanation / Answer
Given that,
Vo = 25 m/s and h = 58.5 meters
(a)The coordinates of initial position of the stone is:
(x0 , y0) = (0,0).
(b) components of the initial velocity is
(Vox, Voy) = (25 m/s, 0 ).
c)first option
d)second option
e)Vx = will be time independent and remains same through out the course of motion.
Vy = Voy + ay t = 0 - 9.8 t = -9.8 t
Vy = -9.8 t
Hence, Vy = -9.8 t
f)X = Vox t + 1/2 ax t2 , but ax = 0
X = 25 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
(g)We have, Y = - 4.9 t2 ( Y = h = 42 m)
-4.9 t2 = - 58.5 => t = 3.46 sec
Hence, t = 3.46 sec
(h)We have, Vx = 25 m/s
and as calculated in the part c and e , we have, Vy = -9.8 t and t = 3.46 sec
Vy = -9.8 x 3.46 = -33.86 m/s
Vf = sqrt [(Vx)2 + (Vy)2] = sqrt [(25)2 + (-33.86)2] = 42.1 m/s
theta = tan -1 ( -34/25) = -53.56 deg below the horizontal
Hence, Vf = 42.1 m/s and theta = -53.56 ° below the horizontal.
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