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 vi = 14.0 m/s. The cliff is h = 50.8 m above a body of water as shown in the figure below.

(a) What are the coordinates of the initial position of the stone? (Hint: Let the student's position be the origin of the coordinate system.) xi = m yi = m

(b) What are the components of the initial velocity of the stone? vix = m/s viy = m/s

(c) What is the appropriate analysis model for the vertical motion of the stone? particle under constant, nonzero acceleration particle under constant speed

(d) What is the appropriate analysis model for the horizontal motion of the stone? particle under constant speed particle under constant, nonzero acceleration

(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: vix, g, and t. Indicate the direction of the velocity with the sign of your answer.) vfx = vfy =

(f) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: vix, g, and t. Indicate the direction of the displacement with the sign of your answer.) xf = yf =

(g) How long after being released does the stone strike the water below the cliff? s

(h) With what speed and angle of impact does the stone land? v = m/s ? = ? counter clockwise from the x-axis

Explanation / Answer

vo=14 m/s==>ht=50.8 m==>vox=vo*cos()=14 m/s==>voy=0 m/s

a) 0i-50.8j

b) vox=14 m/s==>voy=0 m/s

c) g=-9.8 m/s2 acceleration (constant), and constant velocity

d) stone constant velocity in horizontal direction, but no ax!, only ay

in vertical direction

e) flight time=(50.8*2/9.8)3.22 seconds ==>vox=vx=vxfinal=14 m/s for entire

flight because no horizontal acceleration ax, voy=0 m/s because

14 m/s*sin(0o)=0 m/s;vyfinal=voy+ay*t=0+9.8*3.2231.55 m/s absolute

f) vx+vyfinal "(voy+ay*t)"==>range=Vo2*sin(2*)/g

g) look at e

h) v=(142+31.552)34.521 m/s==>=tan(34.5/14)-1= -67.93 degrees

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