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 = 18.2 m/s. The cliff is h = 49.5 m above a body of water as shown in the figure below.

(a) What are the coordinates of the initial position of the stone? (Do not assume that the student is the point of origin.)

(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? (Select all that apply.)

constant velocity motion free fall motion g = 9.8 m/s2 no acceleration from gravity g = 9.8 m/s2

(d) What is the appropriate analysis model for the horizontal motion of the stone? (Select all that apply.)

constant velocity in the horizontal direction g = 9.8 m/s2 g = 9.8 m/s2 free fall motion no acceleration from gravity

(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.)

(f) Write symbolic equations for the position of the stone as a function of time. (Use the following as necessary: vix, g and t.)

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

(h) With what speed and angle of impact does the stone land?

Explanation / Answer

A. Seeing that I cannot see the uploaded image and we're forbidden to put the student at the origin, I'd venture that the coordinates are x = 0m, y = 49.5m

B.

vix = 18.2 m/s

viy = 0

C. For the Y direction, we're looking at something in free fall, where g = -9.8 m/s2

D. For X: Constant velocity, no acceleration due to gravity.

E.

vx = vix (or Vx = 18.2 m/s)

vy = gt

F.

x = vixt ( or x = 18.2m/s . t )

y = 49.5m + gt2 ( or y = 49.5m - 9.8m/s2. t2)

G.

2.247 s.

Found by throwing our numbers into the y equation above, solving for when y = 0, as in:

0 = 45.9m - 9.8m/s2. t2

49.5m = 9.8m/s2. t2

t2 = 49.5m / 9.8m/s2

t = 2.247 s

H.

To find the velocity at impact, use the Pythaghorean Theorem on the two component velocities. Using the number we found in H to determine how long impact takes, we can plug that into the Y velocity equation to get the Y velocity at impact, yielding around -22m/s. Throw that at Pythagoras: v2 = (22m/s)2 + (18.2m/s)2 then solve for v, and we get around 28.57 m/s at impact.

The angle of impact is about 50.43 degrees from the horizontal. This is found by taking the inverse tangent of the two velocities, using the vertical velocity as the opposite angle. Dividing the vertical velocity by the x velocity gets us 1.21, the arctangent of which is 50.43 degrees.

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